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 $(g : X \to Y', s : Y \to Y')$. Then $A = B$ if and only if there exists a morphism $a : Y' \to Y''$ with $a \circ s \in S$ and such that $a \circ f = a \circ g$.

Proof. The equality of $A$ and $B$ means that there exists a commutative diagram

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

with $t \in S$. In particular $u \circ s = v \circ s$. Hence by LMS3 there exists an $s' : Z \to Y''$ in $S$ such that $s' \circ u = s' \circ v$. Setting $a$ equal to this common value does the job. $\square$

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