The Stacks project

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 exists 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. $\square$

Comments (2)

Comment #322 by arp on

Typos: 1. In the sentence, "A simple formal argument, using the properties of E above shows that it suffices to prove "

Comment #323 by arp on

Hmm something went wrong the first time I posted this.


  1. In the sentence, "A simple formal argument, using the properties of E above shows that it suffices to prove ", I think it should say .

  2. In the sentence, "Then we have ," I think the two occurrences of should be replaced with .

  3. In the sentence, "Hence after replacing by , by , and by we may assume that ", I think it should be " by " in place of " by ".

  4. This isn't a typo, but it makes more sense to use instead of in the big diagram at the end of the proof (for associativity of composition).

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