## 4.27 Localization in categories

The basic idea of this section is given a category $\mathcal{C}$ and a set of arrows $S$ to construct a functor $F : \mathcal{C} \to S^{-1}\mathcal{C}$ such that all elements of $S$ become invertible in $S^{-1}\mathcal{C}$ and such that $F$ is universal among all functors with this property. References for this section are [Chapter I, Section 2, GZ] and [Chapter II, Section 2, Verdier].

Definition 4.27.1. Let $\mathcal{C}$ be a category. A set of arrows $S$ of $\mathcal{C}$ is called a left multiplicative system if it has the following properties:

1. The identity of every object of $\mathcal{C}$ is in $S$ and the composition of two composable elements of $S$ is in $S$.

2. Every solid diagram

$\xymatrix{ X \ar[d]_ t \ar[r]_ g & Y \ar@{..>}[d]^ s \\ Z \ar@{..>}[r]^ f & W }$

with $t \in S$ can be completed to a commutative dotted square with $s \in S$.

3. For every pair of morphisms $f, g : X \to Y$ and $t \in S$ with target $X$ such that $f \circ t = g \circ t$ there exists an $s \in S$ with source $Y$ such that $s \circ f = s \circ g$.

A set of arrows $S$ of $\mathcal{C}$ is called a right multiplicative system if it has the following properties:

1. The identity of every object of $\mathcal{C}$ is in $S$ and the composition of two composable elements of $S$ is in $S$.

2. Every solid diagram

$\xymatrix{ X \ar@{..>}[d]_ t \ar@{..>}[r]_ g & Y \ar[d]^ s \\ Z \ar[r]^ f & W }$

with $s \in S$ can be completed to a commutative dotted square with $t \in S$.

3. For every pair of morphisms $f, g : X \to Y$ and $s \in S$ with source $Y$ such that $s \circ f = s \circ g$ there exists a $t \in S$ with target $X$ such that $f \circ t = g \circ t$.

A set of arrows $S$ of $\mathcal{C}$ is called a multiplicative system if it is both a left multiplicative system and a right multiplicative system. In other words, this means that MS1, MS2, MS3 hold, where MS1 $=$ LMS1 $+$ RMS1, MS2 $=$ LMS2 $+$ RMS2, and MS3 $=$ LMS3 $+$ RMS3. (That said, of course LMS1 $=$ RMS1 $=$ MS1.)

These conditions are useful to construct the categories $S^{-1}\mathcal{C}$ as follows.

Left calculus of fractions. Let $\mathcal{C}$ be a category and let $S$ be a left multiplicative system. We define a new category $S^{-1}\mathcal{C}$ as follows (we verify this works in the proof of Lemma 4.27.2):

1. We set $\mathop{\mathrm{Ob}}\nolimits (S^{-1}\mathcal{C}) = \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$.

2. Morphisms $X \to Y$ of $S^{-1}\mathcal{C}$ are given by pairs $(f : X \to Y', s : Y \to Y')$ with $s \in S$ up to equivalence. (The equivalence is defined below. Think of the equivalence class of a pair $(f, s)$ as $s^{-1}f : X \to Y$.)

3. Two pairs $(f_1 : X \to Y_1, s_1 : Y \to Y_1)$ and $(f_2 : X \to Y_2, s_2 : Y \to Y_2)$ are said to be equivalent if there exist a third pair $(f_3 : X \to Y_3, s_3 : Y \to Y_3)$ and morphisms $u : Y_1 \to Y_3$ and $v : Y_2 \to Y_3$ of $\mathcal{C}$ fitting into the commutative diagram

$\xymatrix{ & Y_1 \ar[d]^ u & \\ X \ar[ru]^{f_1} \ar[r]^{f_3} \ar[rd]_{f_2} & Y_3 & Y \ar[lu]_{s_1} \ar[l]_{s_3} \ar[ld]^{s_2} \\ & Y_2 \ar[u]_ v & }$
4. The composition of the equivalence classes of the pairs $(f : X \to Y', s : Y \to Y')$ and $(g : Y \to Z', t : Z \to Z')$ is defined as the equivalence class of a pair $(h \circ f : X \to Z'', u \circ t : Z \to Z'')$ where $h$ and $u \in S$ are chosen to fit into a commutative diagram

$\xymatrix{ Y \ar[d]_ s \ar[r]_ g & Z' \ar[d]^ u \\ Y' \ar[r]^ h & Z'' }$

which exists by assumption.

5. The identity morphism $X \to X$ in $S^{-1} \mathcal{C}$ is the equivalence class of the pair $(\text{id} : X \to X, \text{id} : X \to X)$.

Lemma 4.27.2. Let $\mathcal{C}$ be a category and let $S$ be a left multiplicative system.

1. The relation on pairs defined above is an equivalence relation.

2. The composition rule given above is well defined on equivalence classes.

3. Composition is associative (and the identity morphisms satisfy the identity axioms), and hence $S^{-1}\mathcal{C}$ is a category.

Proof. Proof of (1). Let us say two pairs $p_1 = (f_1 : X \to Y_1, s_1 : Y \to Y_1)$ and $p_2 = (f_2 : X \to Y_2, s_2 : Y \to Y_2)$ are elementary equivalent if there exists a morphism $a : Y_1 \to Y_2$ of $\mathcal{C}$ such that $a \circ f_1 = f_2$ and $a \circ s_1 = s_2$. Diagram:

$\xymatrix{ X \ar@{=}[d] \ar[r]_{f_1} & Y_1 \ar[d]^ a & Y \ar[l]^{s_1} \ar@{=}[d] \\ X \ar[r]^{f_2} & Y_2 & Y \ar[l]_{s_2} }$

Let us denote this property by saying $p_1Ep_2$. Note that $pEp$ and $aEb, bEc \Rightarrow aEc$. (Despite its name, $E$ is not an equivalence relation.) Part (1) claims that the relation $p \sim p' \Leftrightarrow \exists q: pEq \wedge p'Eq$ (where $q$ is supposed to be a pair satisfying the same conditions as $p$ and $p'$) is an equivalence relation. A simple formal argument, using the properties of $E$ above, shows that it suffices to prove $p_3Ep_1, p_3Ep_2 \Rightarrow p_1 \sim p_2$. Thus suppose that we are given a commutative diagram

$\xymatrix{ & Y_1 & \\ X \ar[ru]^{f_1} \ar[r]^{f_3} \ar[rd]_{f_2} & Y_3 \ar[u]_{a_{31}} \ar[d]^{a_{32}} & Y \ar[lu]_{s_1} \ar[l]_{s_3} \ar[ld]^{s_2} \\ & Y_2 & }$

with $s_ i \in S$. First we apply LMS2 to get a commutative diagram

$\xymatrix{ Y \ar[d]_{s_1} \ar[r]_{s_2} & Y_2 \ar@{..>}[d]^{a_{24}} \\ Y_1 \ar@{..>}[r]^{a_{14}} & Y_4 }$

with $a_{24} \in S$. Then, we have

$a_{14} \circ a_{31} \circ s_3 = a_{14} \circ s_1 = a_{24} \circ s_2 = a_{24} \circ a_{32} \circ s_3.$

Hence, by LMS3, there exists a morphism $s_{44} : Y_4 \to Y'_4$ such that $s_{44} \in S$ and $s_{44} \circ a_{14} \circ a_{31} = s_{44} \circ a_{24} \circ a_{32}$. Hence, after replacing $Y_4$, $a_{14}$ and $a_{24}$ by $Y'_4$, $s_{44} \circ a_{14}$ and $s_{44} \circ a_{24}$, we may assume that $a_{14} \circ a_{31} = a_{24} \circ a_{32}$ (and we still have $a_{24} \in S$ and $a_{14} \circ s_1 = a_{24} \circ s_2$). Set

$f_4 = a_{14} \circ f_1 = a_{14} \circ a_{31} \circ f_3 = a_{24} \circ a_{32} \circ f_3 = a_{24} \circ f_2$

and $s_4 = a_{14} \circ s_1 = a_{24} \circ s_2$. Then, the diagram

$\xymatrix{ X \ar@{=}[d] \ar[r]_{f_1} & Y_1 \ar[d]^{a_{14}} & Y \ar[l]^{s_1} \ar@{=}[d] \\ X \ar[r]^{f_4} & Y_4 & Y \ar[l]_{s_4} }$

commutes, and we have $s_4 \in S$ (by LMS1). Thus, $p_1 E p_4$, where $p_4 = (f_4, s_4)$. Similarly, $p_2 E p_4$. Combining these, we find $p_1 \sim p_2$.

Proof of (2). Let $p = (f : X \to Y', s : Y \to Y')$ and $q = (g : Y \to Z', t : Z \to Z')$ be pairs as in the definition of composition above. To compose we choose a diagram

$\xymatrix{ Y \ar[d]_ s \ar[r]_ g & Z' \ar[d]^{u_2} \\ Y' \ar[r]^{h_2} & Z_2 }$

with $u_2 \in S$. We first show that the equivalence class of the pair $r_2 = (h_2 \circ f : X \to Z_2, u_2 \circ t : Z \to Z_2)$ is independent of the choice of $(Z_2, h_2, u_2)$. Namely, suppose that $(Z_3, h_3, u_3)$ is another choice with corresponding composition $r_3 = (h_3 \circ f : X \to Z_3, u_3 \circ t : Z \to Z_3)$. Then by LMS2 we can choose a diagram

$\xymatrix{ Z' \ar[d]_{u_2} \ar[r]_{u_3} & Z_3 \ar[d]^{u_{34}} \\ Z_2 \ar[r]^{h_{24}} & Z_4 }$

with $u_{34} \in S$. We have $h_2 \circ s = u_2 \circ g$ and similarly $h_3 \circ s = u_3 \circ g$. Now,

$u_{34} \circ h_3 \circ s = u_{34} \circ u_3 \circ g = h_{24} \circ u_2 \circ g = h_{24} \circ h_2 \circ s.$

Hence, LMS3 shows that there exist a $Z'_4$ and an $s_{44} : Z_4 \to Z'_4$ such that $s_{44} \circ u_{34} \circ h_3 = s_{44} \circ h_{24} \circ h_2$. Replacing $Z_4$, $h_{24}$ and $u_{34}$ by $Z'_4$, $s_{44} \circ h_{24}$ and $s_{44} \circ u_{34}$, we may assume that $u_{34} \circ h_3 = h_{24} \circ h_2$. Meanwhile, the relations $u_{34} \circ u_3 = h_{24} \circ u_2$ and $u_{34} \in S$ continue to hold. We can now set $h_4 = u_{34} \circ h_3 = h_{24} \circ h_2$ and $u_4 = u_{34} \circ u_3 = h_{24} \circ u_2$. Then, we have a commutative diagram

$\xymatrix{ X \ar@{=}[d] \ar[r]_{h_2\circ f} & Z_2 \ar[d]^{h_{24}} & Z \ar[l]^{u_2 \circ t} \ar@{=}[d] \\ X \ar@{=}[d] \ar[r]^{h_4\circ f} & Z_4 & Z \ar@{=}[d] \ar[l]_{u_4 \circ t} \\ X \ar[r]^{h_3 \circ f} & Z_3 \ar[u]^{u_{34}} & Z \ar[l]_{u_3 \circ t} }$

Hence we obtain a pair $r_4 = (h_4 \circ f : X \to Z_4, u_4 \circ t : Z \to Z_4)$ and the above diagram shows that we have $r_2Er_4$ and $r_3Er_4$, whence $r_2 \sim r_3$, as desired. Thus it now makes sense to define $p \circ q$ as the equivalence class of all possible pairs $r$ obtained as above.

To finish the proof of (2) we have to show that given pairs $p_1, p_2, q$ such that $p_1Ep_2$ then $p_1 \circ q = p_2 \circ q$ and $q \circ p_1 = q \circ p_2$ whenever the compositions make sense. To do this, write $p_1 = (f_1 : X \to Y_1, s_1 : Y \to Y_1)$ and $p_2 = (f_2 : X \to Y_2, s_2 : Y \to Y_2)$ and let $a : Y_1 \to Y_2$ be a morphism of $\mathcal{C}$ such that $f_2 = a \circ f_1$ and $s_2 = a \circ s_1$. First assume that $q = (g : Y \to Z', t : Z \to Z')$. In this case choose a commutative diagram as the one on the left

$\vcenter { \xymatrix{ Y \ar[d]_{s_2} \ar[r]^ g & Z' \ar[d]^ u \\ Y_2 \ar[r]^ h & Z'' } } \quad \Rightarrow \quad \vcenter { \xymatrix{ Y \ar[d]_{s_1} \ar[r]^ g & Z' \ar[d]^ u \\ Y_1 \ar[r]^{h \circ a} & Z'' } }$

(with $u \in S$), which implies the diagram on the right is commutative as well. Using these diagrams we see that both compositions $q \circ p_1$ and $q \circ p_2$ are the equivalence class of $(h \circ a \circ f_1 : X \to Z'', u \circ t : Z \to Z'')$. Thus $q \circ p_1 = q \circ p_2$. The proof of the other case, in which we have to show $p_1 \circ q = p_2 \circ q$, is omitted. (It is similar to the case we did.)

Proof of (3). We have to prove associativity of composition. Consider a solid diagram

$\xymatrix{ & & & Z \ar[d] \\ & & Y \ar[d] \ar[r] & Z' \ar@{..>}[d] \\ & X \ar[d] \ar[r] & Y' \ar@{..>}[d] \ar@{..>}[r] & Z'' \ar@{..>}[d] \\ W \ar[r] & X' \ar@{..>}[r] & Y'' \ar@{..>}[r] & Z''' }$

(whose vertical arrows belong to $S$) which gives rise to three composable pairs. Using LMS2 we can choose the dotted arrows making the squares commutative and such that the vertical arrows are in $S$. Then it is clear that the composition of the three pairs is the equivalence class of the pair $(W \to Z''', Z \to Z''')$ gotten by composing the horizontal arrows on the bottom row and the vertical arrows on the right column.

We leave it to the reader to check the identity axioms. $\square$

Remark 4.27.3. The motivation for the construction of $S^{-1} \mathcal{C}$ is to “force” the morphisms in $S$ to be invertible by artificially creating inverses to them (at the cost of some existing morphisms possibly becoming identified with each other). This is similar to the localization of a commutative ring at a multiplicative subset, and more generally to the localization of a noncommutative ring at a right denominator set (see [Section 10A, Lam]). This is more than just a similarity: The construction of $S^{-1} \mathcal{C}$ (or, more precisely, its version for additive categories $\mathcal{C}$) actually generalizes the latter type of localization. Namely, a noncommutative ring can be viewed as a pre-additive category with a single object (the morphisms being the elements of the ring); a multiplicative subset of this ring then becomes a set $S$ of morphisms satisfying LMS1 (aka RMS1). Then, the conditions RMS2 and RMS3 for this category and this subset $S$ translate into the two conditions (“right permutable” and “right reversible”) of a right denominator set (and similarly for LMS and left denominator sets), and $S^{-1} \mathcal{C}$ (with a properly defined additive structure) is the one-object category corresponding to the localization of the ring.

Definition 4.27.4. Let $\mathcal{C}$ be a category and let $S$ be a left multiplicative system of morphisms of $\mathcal{C}$. Given any morphism $f : X \to Y'$ in $\mathcal{C}$ and any morphism $s : Y \to Y'$ in $S$, we denote by $s^{-1} f$ the equivalence class of the pair $(f : X \to Y', s : Y \to Y')$. This is a morphism from $X$ to $Y$ in $S^{-1} \mathcal{C}$.

This notation is suggestive, and the things it suggests are true: Given any morphism $f : X \to Y'$ in $\mathcal{C}$ and any two morphisms $s : Y \to Y'$ and $t : Y' \to Y''$ in $S$, we have $\left(t \circ s\right)^{-1} \left(t \circ f\right) = s^{-1} f$. Also, for any $f : X \to Y'$ and $g : Y' \to Z'$ in $\mathcal{C}$ and all $s : Z \to Z'$ in $S$, we have $s^{-1} \left(g \circ f\right) = \left(s^{-1} g\right) \circ \left(\text{id}_{Y'}^{-1} f\right)$. Finally, for any $f : X \to Y'$ in $\mathcal{C}$, all $s : Y \to Y'$ in $S$, and $t : Z \to Y$ in $S$, we have $\left(s \circ t\right)^{-1} f = \left(t^{-1} \text{id}_ Y\right) \circ \left(s^{-1} f\right)$. This is all clear from the definition. We can “write any finite collection of morphisms with the same target as fractions with common denominator”.

Lemma 4.27.5. Let $\mathcal{C}$ be a category and let $S$ be a left multiplicative system of morphisms of $\mathcal{C}$. Given any finite collection $g_ i : X_ i \to Y$ of morphisms of $S^{-1}\mathcal{C}$ (indexed by $i$), we can find an element $s : Y \to Y'$ of $S$ and a family of morphisms $f_ i : X_ i \to Y'$ of $\mathcal{C}$ such that each $g_ i$ is the equivalence class of the pair $(f_ i : X_ i \to Y', s : Y \to Y')$.

Proof. For each $i$ choose a representative $(X_ i \to Y_ i, s_ i : Y \to Y_ i)$ of $g_ i$. The lemma follows if we can find a morphism $s : Y \to Y'$ in $S$ such that for each $i$ there is a morphism $a_ i : Y_ i \to Y'$ with $a_ i \circ s_ i = s$. If we have two indices $i = 1, 2$, then we can do this by completing the square

$\xymatrix{ Y \ar[d]_{s_1} \ar[r]_{s_2} & Y_2 \ar[d]^{t_2} \\ Y_1 \ar[r]^{a_1} & Y' }$

with $t_2 \in S$ as is possible by Definition 4.27.1. Then $s = t_2 \circ s_2 \in S$ works. If we have $n > 2$ morphisms, then we use the above trick to reduce to the case of $n - 1$ morphisms, and we win by induction. $\square$

There is an easy characterization of equality of morphisms if they have the same denominator.

Lemma 4.27.6. Let $\mathcal{C}$ be a category and let $S$ be a left multiplicative system of morphisms of $\mathcal{C}$. Let $A, B : X \to Y$ be morphisms of $S^{-1}\mathcal{C}$ which are the equivalence classes of $(f : X \to Y', s : Y \to Y')$ and $(g : X \to Y', s : Y \to Y')$. The following are equivalent

1. $A = B$

2. there exists a morphism $t : Y' \to Y''$ in $S$ with $t \circ f = t \circ g$, and

3. there exists a morphism $a : Y' \to Y''$ such that $a \circ f = a \circ g$ and $a \circ s \in S$.

Proof. We are going to use that $S^{-1}\mathcal{C}$ is a category (Lemma 4.27.2) and we will use the notation of Definition 4.27.4 as well as the discussion following that definition to identify some morphisms in $S^{-1}\mathcal{C}$. Thus we write $A = s^{-1}f$ and $B = s^{-1}g$.

If $A = B$ then $(\text{id}_{Y'}^{-1}s) \circ A = (\text{id}_{Y'}^{-1}s) \circ B$. We have $(\text{id}_{Y'}^{-1}s) \circ A = \text{id}_{Y'}^{-1}f$ and $(\text{id}_{Y'}^{-1}s) \circ B = \text{id}_{Y'}^{-1}g$. The equality of $\text{id}_{Y'}^{-1}f$ and $\text{id}_{Y'}^{-1}g$ means by definition that there exists a commutative diagram

$\xymatrix{ & Y' \ar[d]^ u & \\ X \ar[ru]^ f \ar[r]^ h \ar[rd]_ g & Z & Y' \ar[lu]_{\text{id}_{Y'}} \ar[l]_ t \ar[ld]^{\text{id}_{Y'}} \\ & Y' \ar[u]_ v & }$

with $t \in S$. In particular $u = v = t \in S$ and $t \circ f = t\circ g$. Thus (1) implies (2).

The implication (2) $\Rightarrow$ (3) is immediate. Assume $a$ is as in (3). Denote $s' = a \circ s \in S$. Then $\text{id}_{Y''}^{-1}s'$ is an isomorphism in the category $S^{-1}\mathcal{C}$ (with inverse $(s')^{-1}\text{id}_{Y''}$). Thus to check $A = B$ it suffices to check that $\text{id}_{Y''}^{-1}s' \circ A = \text{id}_{Y''}^{-1}s' \circ B$. We compute using the rules discussed in the text following Definition 4.27.4 that $\text{id}_{Y''}^{-1}s' \circ A = \text{id}_{Y''}^{-1}(a \circ s) \circ s^{-1}f = \text{id}_{Y''}^{-1}(a \circ f) = \text{id}_{Y''}^{-1}(a \circ g) = \text{id}_{Y''}^{-1}(a \circ s) \circ s^{-1}g = \text{id}_{Y''}^{-1}s' \circ B$ and we see that (1) is true. $\square$

Remark 4.27.7. Let $\mathcal{C}$ be a category. Let $S$ be a left multiplicative system. Given an object $Y$ of $\mathcal{C}$ we denote $Y/S$ the category whose objects are $s : Y \to Y'$ with $s \in S$ and whose morphisms are commutative diagrams

$\xymatrix{ & Y \ar[ld]_ s \ar[rd]^ t & \\ Y' \ar[rr]^ a & & Y'' }$

where $a : Y' \to Y''$ is arbitrary. We claim that the category $Y/S$ is filtered (see Definition 4.19.1). Namely, LMS1 implies that $\text{id}_ Y : Y \to Y$ is in $Y/S$; hence $Y/S$ is nonempty. LMS2 implies that given $s_1 : Y \to Y_1$ and $s_2 : Y \to Y_2$ we can find a diagram

$\xymatrix{ Y \ar[d]_{s_1} \ar[r]_{s_2} & Y_2 \ar[d]^ t \\ Y_1 \ar[r]^ a & Y_3 }$

with $t \in S$. Hence $s_1 : Y \to Y_1$ and $s_2 : Y \to Y_2$ both have maps to $t \circ s_2 : Y \to Y_3$ in $Y/S$. Finally, given two morphisms $a, b$ from $s_1 : Y \to Y_1$ to $s_2 : Y \to Y_2$ in $Y/S$ we see that $a \circ s_1 = b \circ s_1$; hence by LMS3 there exists a $t : Y_2 \to Y_3$ in $S$ such that $t \circ a = t \circ b$. Now the combined results of Lemmas 4.27.5 and 4.27.6 tell us that

4.27.7.1
\begin{equation} \label{categories-equation-left-localization-morphisms-colimit} \mathop{\mathrm{Mor}}\nolimits _{S^{-1}\mathcal{C}}(X, Y) = \mathop{\mathrm{colim}}\nolimits _{(s : Y \to Y') \in Y/S} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(X, Y') \end{equation}

This formula expressing morphism sets in $S^{-1}\mathcal{C}$ as a filtered colimit of morphism sets in $\mathcal{C}$ is occasionally useful.

Lemma 4.27.8. Let $\mathcal{C}$ be a category and let $S$ be a left multiplicative system of morphisms of $\mathcal{C}$.

1. The rules $X \mapsto X$ and $(f : X \to Y) \mapsto (f : X \to Y, \text{id}_ Y : Y \to Y)$ define a functor $Q : \mathcal{C} \to S^{-1}\mathcal{C}$.

2. For any $s \in S$ the morphism $Q(s)$ is an isomorphism in $S^{-1}\mathcal{C}$.

3. If $G : \mathcal{C} \to \mathcal{D}$ is any functor such that $G(s)$ is invertible for every $s \in S$, then there exists a unique functor $H : S^{-1}\mathcal{C} \to \mathcal{D}$ such that $H \circ Q = G$.

Proof. Parts (1) and (2) are clear. (In (2), the inverse of $Q(s)$ is the equivalence class of the pair $(\text{id}_ Y, s)$.) To see (3) just set $H(X) = G(X)$ and set $H((f : X \to Y', s : Y \to Y')) = G(s)^{-1} \circ G(f)$. Details omitted. $\square$

Lemma 4.27.9. Let $\mathcal{C}$ be a category and let $S$ be a left multiplicative system of morphisms of $\mathcal{C}$. The localization functor $Q : \mathcal{C} \to S^{-1}\mathcal{C}$ commutes with finite colimits.

Proof. Let $\mathcal{I}$ be a finite category and let $\mathcal{I} \to \mathcal{C}$, $i \mapsto X_ i$ be a functor whose colimit exists. Then using (4.27.7.1), the fact that $Y/S$ is filtered, and Lemma 4.19.2 we have

\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{S^{-1}\mathcal{C}}(Q(\mathop{\mathrm{colim}}\nolimits X_ i), Q(Y)) & = \mathop{\mathrm{colim}}\nolimits _{(s : Y \to Y') \in Y/S} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(\mathop{\mathrm{colim}}\nolimits X_ i, Y') \\ & = \mathop{\mathrm{colim}}\nolimits _{(s : Y \to Y') \in Y/S} \mathop{\mathrm{lim}}\nolimits _ i \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(X_ i, Y') \\ & = \mathop{\mathrm{lim}}\nolimits _ i \mathop{\mathrm{colim}}\nolimits _{(s : Y \to Y') \in Y/S} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(X_ i, Y') \\ & = \mathop{\mathrm{lim}}\nolimits _ i \mathop{\mathrm{Mor}}\nolimits _{S^{-1}\mathcal{C}}(Q(X_ i), Q(Y)) \end{align*}

and this isomorphism commutes with the projections from both sides to the set $\mathop{\mathrm{Mor}}\nolimits _{S^{-1}\mathcal{C}}(Q(X_ j), Q(Y))$ for each $j \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$. Thus, $Q(\mathop{\mathrm{colim}}\nolimits X_ i)$ satisfies the universal property for the colimit of the functor $i \mapsto Q(X_ i)$; hence, it is this colimit, as desired. $\square$

Lemma 4.27.10. Let $\mathcal{C}$ be a category. Let $S$ be a left multiplicative system. If $f : X \to Y$, $f' : X' \to Y'$ are two morphisms of $\mathcal{C}$ and if

$\xymatrix{ Q(X) \ar[d]_{Q(f)} \ar[r]_ a & Q(X') \ar[d]^{Q(f')} \\ Q(Y) \ar[r]^ b & Q(Y') }$

is a commutative diagram in $S^{-1}\mathcal{C}$, then there exist a morphism $f'' : X'' \to Y''$ in $\mathcal{C}$ and a commutative diagram

$\xymatrix{ X \ar[d]_ f \ar[r]_ g & X'' \ar[d]^{f''} & X' \ar[d]^{f'} \ar[l]^ s \\ Y \ar[r]^ h & Y'' & Y' \ar[l]_ t }$

in $\mathcal{C}$ with $s, t \in S$ and $a = s^{-1}g$, $b = t^{-1}h$.

Proof. We choose maps and objects in the following way: First write $a = s^{-1}g$ for some $s : X' \to X''$ in $S$ and $g : X \to X''$. By LMS2 we can find $t : Y' \to Y''$ in $S$ and $f'' : X'' \to Y''$ such that

$\xymatrix{ X \ar[d]_ f \ar[r]_ g & X'' \ar[d]^{f''} & X' \ar[d]^{f'} \ar[l]^ s \\ Y & Y'' & Y' \ar[l]_ t }$

commutes. Now in this diagram we are going to repeatedly change our choice of

$X'' \xrightarrow {f''} Y'' \xleftarrow {t} Y'$

by postcomposing both $t$ and $f''$ by a morphism $d : Y'' \to Y'''$ with the property that $d \circ t \in S$. According to Remark 4.27.7 we may after such a replacement assume that there exists a morphism $h : Y \to Y''$ such that $b = t^{-1}h$ holds1. At this point we have everything as in the lemma except that we don't know that the left square of the diagram commutes. But the definition of composition in $S^{-1} \mathcal{C}$ shows that $b \circ Q\left(f\right)$ is the equivalence class of the pair $(h \circ f : X \to Y'', t : Y' \to Y'')$ (since $b$ is the equivalence class of the pair $(h : Y \to Y'', t : Y' \to Y'')$, while $Q\left(f\right)$ is the equivalence class of the pair $(f : X \to Y, \text{id} : Y \to Y)$), while $Q\left(f'\right) \circ a$ is the equivalence class of the pair $(f'' \circ g : X \to Y'', t : Y' \to Y'')$ (since $a$ is the equivalence class of the pair $(g : X \to X'', s : X' \to X'')$, while $Q\left(f'\right)$ is the equivalence class of the pair $(f' : X' \to Y', \text{id} : Y' \to Y')$). Since we know that $b \circ Q\left(f\right) = Q\left(f'\right) \circ a$, we thus conclude that the equivalence classes of the pairs $(h \circ f : X \to Y'', t : Y' \to Y'')$ and $(f'' \circ g : X \to Y'', t : Y' \to Y'')$ are equal. Hence using Lemma 4.27.6 we can find a morphism $d : Y'' \to Y'''$ such that $d \circ t \in S$ and $d \circ h \circ f = d \circ f'' \circ g$. Hence we make one more replacement of the kind described above and we win. $\square$

Right calculus of fractions. Let $\mathcal{C}$ be a category and let $S$ be a right multiplicative system. We define a new category $S^{-1}\mathcal{C}$ as follows (we verify this works in the proof of Lemma 4.27.11):

1. We set $\mathop{\mathrm{Ob}}\nolimits (S^{-1}\mathcal{C}) = \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$.

2. Morphisms $X \to Y$ of $S^{-1}\mathcal{C}$ are given by pairs $(f : X' \to Y, s : X' \to X)$ with $s \in S$ up to equivalence. (The equivalence is defined below. Think of the equivalence class of a pair $(f, s)$ as $fs^{-1} : X \to Y$.)

3. Two pairs $(f_1 : X_1 \to Y, s_1 : X_1 \to X)$ and $(f_2 : X_2 \to Y, s_2 : X_2 \to X)$ are said to be equivalent if there exist a third pair $(f_3 : X_3 \to Y, s_3 : X_3 \to X)$ and morphisms $u : X_3 \to X_1$ and $v : X_3 \to X_2$ of $\mathcal{C}$ fitting into the commutative diagram

$\xymatrix{ & X_1 \ar[ld]_{s_1} \ar[rd]^{f_1} & \\ X & X_3 \ar[l]_{s_3} \ar[u]_ u \ar[d]^ v \ar[r]^{f_3} & Y \\ & X_2 \ar[lu]^{s_2} \ar[ru]_{f_2} & }$
4. The composition of the equivalence classes of the pairs $(f : X' \to Y, s : X' \to X)$ and $(g : Y' \to Z, t : Y' \to Y)$ is defined as the equivalence class of a pair $(g \circ h : X'' \to Z, s \circ u : X'' \to X)$ where $h$ and $u \in S$ are chosen to fit into a commutative diagram

$\xymatrix{ X'' \ar[d]_ u \ar[r]^ h & Y' \ar[d]^ t \\ X' \ar[r]^ f & Y }$

which exists by assumption.

5. The identity morphism $X \to X$ in $S^{-1} \mathcal{C}$ is the equivalence class of the pair $(\text{id} : X \to X, \text{id} : X \to X)$.

Lemma 4.27.11. Let $\mathcal{C}$ be a category and let $S$ be a right multiplicative system.

1. The relation on pairs defined above is an equivalence relation.

2. The composition rule given above is well defined on equivalence classes.

3. Composition is associative (and the identity morphisms satisfy the identity axioms), and hence $S^{-1}\mathcal{C}$ is a category.

Proof. This lemma is dual to Lemma 4.27.2. It follows formally from that lemma by replacing $\mathcal{C}$ by its opposite category in which $S$ is a left multiplicative system. $\square$

Definition 4.27.12. Let $\mathcal{C}$ be a category and let $S$ be a right multiplicative system of morphisms of $\mathcal{C}$. Given any morphism $f : X' \to Y$ in $\mathcal{C}$ and any morphism $s : X' \to X$ in $S$, we denote by $f s^{-1}$ the equivalence class of the pair $(f : X' \to Y, s : X' \to X)$. This is a morphism from $X$ to $Y$ in $S^{-1} \mathcal{C}$.

Identities similar (actually, dual) to the ones in Definition 4.27.4 hold. We can “write any finite collection of morphisms with the same source as fractions with common denominator”.

Lemma 4.27.13. Let $\mathcal{C}$ be a category and let $S$ be a right multiplicative system of morphisms of $\mathcal{C}$. Given any finite collection $g_ i : X \to Y_ i$ of morphisms of $S^{-1}\mathcal{C}$ (indexed by $i$), we can find an element $s : X' \to X$ of $S$ and a family of morphisms $f_ i : X' \to Y_ i$ of $\mathcal{C}$ such that $g_ i$ is the equivalence class of the pair $(f_ i : X' \to Y_ i, s : X' \to X)$.

Proof. This lemma is the dual of Lemma 4.27.5 and follows formally from that lemma by replacing all categories in sight by their opposites. $\square$

There is an easy characterization of equality of morphisms if they have the same denominator.

Lemma 4.27.14. Let $\mathcal{C}$ be a category and let $S$ be a right multiplicative system of morphisms of $\mathcal{C}$. Let $A, B : X \to Y$ be morphisms of $S^{-1}\mathcal{C}$ which are the equivalence classes of $(f : X' \to Y, s : X' \to X)$ and $(g : X' \to Y, s : X' \to X)$. The following are equivalent

1. $A = B$,

2. there exists a morphism $t : X'' \to X'$ in $S$ with $f \circ t = g \circ t$, and

3. there exists a morphism $a : X'' \to X'$ with $f \circ a = g \circ a$ and $s \circ a \in S$.

Proof. This is dual to Lemma 4.27.6. $\square$

Remark 4.27.15. Let $\mathcal{C}$ be a category. Let $S$ be a right multiplicative system. Given an object $X$ of $\mathcal{C}$ we denote $S/X$ the category whose objects are $s : X' \to X$ with $s \in S$ and whose morphisms are commutative diagrams

$\xymatrix{ X' \ar[rd]_ s \ar[rr]_ a & & X'' \ar[ld]^ t \\ & X }$

where $a : X' \to X''$ is arbitrary. The category $S/X$ is cofiltered (see Definition 4.20.1). (This is dual to the corresponding statement in Remark 4.27.7.) Now the combined results of Lemmas 4.27.13 and 4.27.14 tell us that

4.27.15.1
\begin{equation} \label{categories-equation-right-localization-morphisms-colimit} \mathop{\mathrm{Mor}}\nolimits _{S^{-1}\mathcal{C}}(X, Y) = \mathop{\mathrm{colim}}\nolimits _{(s : X' \to X) \in (S/X)^{opp}} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(X', Y) \end{equation}

This formula expressing morphisms in $S^{-1}\mathcal{C}$ as a filtered colimit of morphisms in $\mathcal{C}$ is occasionally useful.

Lemma 4.27.16. Let $\mathcal{C}$ be a category and let $S$ be a right multiplicative system of morphisms of $\mathcal{C}$.

1. The rules $X \mapsto X$ and $(f : X \to Y) \mapsto (f : X \to Y, \text{id}_ X : X \to X)$ define a functor $Q : \mathcal{C} \to S^{-1}\mathcal{C}$.

2. For any $s \in S$ the morphism $Q(s)$ is an isomorphism in $S^{-1}\mathcal{C}$.

3. If $G : \mathcal{C} \to \mathcal{D}$ is any functor such that $G(s)$ is invertible for every $s \in S$, then there exists a unique functor $H : S^{-1}\mathcal{C} \to \mathcal{D}$ such that $H \circ Q = G$.

Proof. This lemma is the dual of Lemma 4.27.8 and follows formally from that lemma by replacing all categories in sight by their opposites. $\square$

Lemma 4.27.17. Let $\mathcal{C}$ be a category and let $S$ be a right multiplicative system of morphisms of $\mathcal{C}$. The localization functor $Q : \mathcal{C} \to S^{-1}\mathcal{C}$ commutes with finite limits.

Proof. This is dual to Lemma 4.27.9. $\square$

Lemma 4.27.18. Let $\mathcal{C}$ be a category. Let $S$ be a right multiplicative system. If $f : X \to Y$, $f' : X' \to Y'$ are two morphisms of $\mathcal{C}$ and if

$\xymatrix{ Q(X) \ar[d]_{Q(f)} \ar[r]_ a & Q(X') \ar[d]^{Q(f')} \\ Q(Y) \ar[r]^ b & Q(Y') }$

is a commutative diagram in $S^{-1}\mathcal{C}$, then there exist a morphism $f'' : X'' \to Y''$ in $\mathcal{C}$ and a commutative diagram

$\xymatrix{ X \ar[d]_ f & X'' \ar[l]^ s \ar[d]^{f''} \ar[r]_ g & X' \ar[d]^{f'} \\ Y & Y'' \ar[l]_ t \ar[r]^ h & Y' }$

in $\mathcal{C}$ with $s, t \in S$ and $a = gs^{-1}$, $b = ht^{-1}$.

Proof. This lemma is dual to Lemma 4.27.10. $\square$

Multiplicative systems and two sided calculus of fractions. If $S$ is a multiplicative system then left and right calculus of fractions give canonically isomorphic categories.

Lemma 4.27.19. Let $\mathcal{C}$ be a category and let $S$ be a multiplicative system. The category of left fractions and the category of right fractions $S^{-1}\mathcal{C}$ are canonically isomorphic.

Proof. Denote $\mathcal{C}_{left}$, $\mathcal{C}_{right}$ the two categories of fractions. By the universal properties of Lemmas 4.27.8 and 4.27.16 we obtain functors $\mathcal{C}_{left} \to \mathcal{C}_{right}$ and $\mathcal{C}_{right} \to \mathcal{C}_{left}$. By the uniqueness statement in the universal properties, these functors are each other's inverse. $\square$

Definition 4.27.20. Let $\mathcal{C}$ be a category and let $S$ be a multiplicative system. We say $S$ is saturated if, in addition to MS1, MS2, MS3, we also have

1. Given three composable morphisms $f, g, h$, if $fg, gh \in S$, then $g \in S$.

Note that a saturated multiplicative system contains all isomorphisms. Moreover, if $f, g, h$ are composable morphisms in a category and $fg, gh$ are isomorphisms, then $g$ is an isomorphism (because then $g$ has both a left and a right inverse, hence is invertible).

Lemma 4.27.21. Let $\mathcal{C}$ be a category and let $S$ be a multiplicative system. Denote $Q : \mathcal{C} \to S^{-1}\mathcal{C}$ the localization functor. The set

$\hat S = \{ f \in \text{Arrows}(\mathcal{C}) \mid Q(f) \text{ is an isomorphism}\}$

is equal to

$S' = \{ f \in \text{Arrows}(\mathcal{C}) \mid \text{there exist }g, h\text{ such that }gf, fh \in S\}$

and is the smallest saturated multiplicative system containing $S$. In particular, if $S$ is saturated, then $\hat S = S$.

Proof. It is clear that $S \subset S' \subset \hat S$ because elements of $S'$ map to morphisms in $S^{-1}\mathcal{C}$ which have both left and right inverses. Note that $S'$ satisfies MS4, and that $\hat S$ satisfies MS1. Next, we prove that $S' = \hat S$.

Let $f \in \hat S$. Let $s^{-1}g = ht^{-1}$ be the inverse morphism in $S^{-1}\mathcal{C}$. (We may use both left fractions and right fractions to describe morphisms in $S^{-1}\mathcal{C}$, see Lemma 4.27.19.) The relation $\text{id}_ X = s^{-1}gf$ in $S^{-1}\mathcal{C}$ means there exists a commutative diagram

$\xymatrix{ & X' \ar[d]^ u & \\ X \ar[ru]^{gf} \ar[r]^{f'} \ar[rd]_{\text{id}_ X} & X'' & X \ar[lu]_ s \ar[l]_{s'} \ar[ld]^{\text{id}_ X} \\ & X \ar[u]_ v & }$

for some morphisms $f', u, v$ and $s' \in S$. Hence $ugf = s' \in S$. Similarly, using that $\text{id}_ Y = fht^{-1}$ one proves that $fhw \in S$ for some $w$. We conclude that $f \in S'$. Thus $S' = \hat S$. Provided we prove that $S' = \hat S$ is a multiplicative system it is now clear that this implies that $S' = \hat S$ is the smallest saturated system containing $S$.

Our remarks above take care of MS1 and MS4, so to finish the proof of the lemma we have to show that LMS2, RMS2, LMS3, RMS3 hold for $\hat S$. Let us check that LMS2 holds for $\hat S$. Suppose we have a solid diagram

$\xymatrix{ X \ar[d]_ t \ar[r]_ g & Y \ar@{..>}[d]^ s \\ Z \ar@{..>}[r]^ f & W }$

with $t \in \hat S$. Pick a morphism $a : Z \to Z'$ such that $at \in S$. Then we can use LMS2 for $S$ to find a commutative diagram

$\xymatrix{ X \ar[d]_ t \ar[r]_ g & Y \ar[dd]^ s \\ Z \ar[d]_ a \\ Z' \ar[r]^{f'} & W }$

and setting $f = f' \circ a$ we win. The proof of RMS2 is dual to this. Finally, suppose given a pair of morphisms $f, g : X \to Y$ and $t \in \hat S$ with target $X$ such that $ft = gt$. Then we pick a morphism $b$ such that $tb \in S$. Then $ftb = gtb$ which implies by LMS3 for $S$ that there exists an $s \in S$ with source $Y$ such that $sf = sg$ as desired. The proof of RMS3 is dual to this. $\square$

 Here is a more down-to-earth way to see this: Write $b = q^{-1}i$ for some $q : Y' \to Z$ in $S$ and some $i : Y \to Z$. By LMS2 we can find $r : Y'' \to Y'''$ in $S$ and $j : Z \to Y'''$ such that $j \circ q = r \circ t$. Now, set $d = r$ and $h = j \circ i$.

Comment #1318 by jojo on

In the phrase just above lemma $4.26.16$, " ... calculus of fractions given canonically ..." should be " ... calculus of fractions give canonically ..."

Comment #1913 by luke on

In the Remark 4.26.3, "... can be viewed as an additive category ...", should be "preadditive" since there is no zero object in this category.

Comment #3376 by Kazuki Masugi on

There are typos in the proof of the Lemma 4.26.10; b is equivalence class of the pair (h , t) and a is of (g , s).

Comment #3976 by qiao on

in Lemma 05Q2. the proof is saying " finite limit commute with Q " while the statement of the lemma is " commute with finite colimit "

Comment #4104 by on

@#3976. Dear qiao, I do not understand your comment. I carefully checked Lemma 4.27.9 and its proof and everything seems fine to me.

Comment #4241 by Si Yu How on

Why isn't there a mention of the (optional) locally small condition of a (left, right) multiplicative system as in Weibel's homological algebra book? Without this condition $S^{-1}\mathcal{C}$ may not be locally small even though $\mathcal{C}$ is.

Comment #4417 by on

In the stacks project all categories have sets of objects and sets of morphisms unless otherwise mentioned. See Remark 4.2.2. The list in this remark needs to be updated however and overall some slippage has crept in.

Comment #5010 by JoseTomas on

I think Lemma 4.26.6 can be simplified (even improved): $A=B \iff s^{-1}f=s^{-1}g \iff f=g$ where equalities are in the category $S^{-1}C$ and we are left composing with $s$ in this category (use of functor $Q$ is implied).

And $f=g$ in $S^{-1}C$ if, and only if, exists $t \in S$ such that $tf=tg$ where the equality is now in $C$. So I think, in reality, condition does not depends on $s$

Comment #5248 by on

@#5010: Please leave a detailed comment on a particular lemma on the page of that lemma. This will really help to have a focused discussion of the particulars. Unfortunately, I did not understand your comment, or put differently I think it doesn't work. Can you explain again please?

Comment #6709 by JoseTomas on

@#5010. There was a mistake in lemma number. I was refering to Lemma 04VF (4.27.6)

$A=B \iff Q(s)^{-1} \circ Q(f)= Q(s)^{-1} \circ Q(g) \iff Q(f)=Q(g)$, where $Q$ is localization functor

And $Q(f)=Q(g)$ if, and only if, exists $t \in S$ such that $tf=tg$ (it is a particular result of the lemma for $s=id$)

Comment #6910 by on

OK, I now looked again at what you said. First of all, we cannot use $Q$ in this lemma because the functor $Q$ is introduced later. But also, I think your statement is incorrect. Because the particular case of the lemma for $s = \text{id}$ as you formulated it doesn't hold. Namely, given a morphisms $a : Y' \to Y''$ and $s : Y \to Y'$ such that $s$ and $a \circ s$ are in $S$, then it isn't necessarily the case that $a$ is in $S$. But worse: there may not be any $t : Y'' \to Y''$ in $S$ such that $t \circ a$ is in $S$. This means that what you said cannot work (I am too lazy to make an explicit counter example -- sorry). If the left multiplicative system is a saturated multiplicative system (Definition 4.27.20), then it does hold (because then $a$ is forced to be in $S$).

Comment #7151 by Jose Tomas on

Hello Johan.

Regarding use of functor $Q$: Leave lemma 04VF (4.27.6) in its original version and let prove a "refined" version at the end of the section. Then you can use $Q$ and you can also use the "base" version.

Let apply original version to $C=(f:X \to Y', id_{Y'}:Y' \to Y')$ and $D=(g:X \to Y', id_{Y'}:Y' \to Y')$. $id_{Y'} \in S$ by property $LMS1$ of left multiplicative systems.

Applying lemma 04VF (4.27.6), $Q(f)=Q(g) \iff C=D \iff \exists a:Y' \to Y''$ with $a \in S$ and such that $a \circ f=a \circ g". This proves the second line of my original statement. First line remains valid. Comment #7152 by on Hahaha, I finally get it! Thanks for getting back to me! Let me contemplate adding this for when I next go through all the comments. Comment #7298 by on OK, now finally fixed this. See this commit. If you want to have your name listed on the contributors list, then please somehow communicate your full name to me. ## Post a comment Your email address will not be published. Required fields are marked. In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi\$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

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