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. $\square$

 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 #326 by arp on

Typos:

1. In the first line of the proof, it should say $g: X \rightarrow X''$ not $h: X \rightarrow X''$.

2. In the second to last line of the proof, the morphism $d$ should be $d: Y'' \rightarrow Y'''$ not $d: X''' \rightarrow X''$.

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