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

$A = B$

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

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.
$\square$

## Comments (1)

Comment #8334 by ElĂas Guisado on

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