The Stacks project

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:

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

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

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

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

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

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

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

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,

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

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

(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

(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. $\square$

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

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. $\square$

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

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. $\square$

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. $\square$

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

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. $\square$

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

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

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. $\square$

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. $\square$

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. $\square$

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

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. $\square$

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

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

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. $\square$

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

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

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

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. $\square$