Lemma 4.27.5. Let $\mathcal{C}$ be a category and let $S$ be a left multiplicative system of morphisms of $\mathcal{C}$. Given any finite collection $g_ i : X_ i \to Y$ of morphisms of $S^{-1}\mathcal{C}$ (indexed by $i$), we can find an element $s : Y \to Y'$ of $S$ and a family of morphisms $f_ i : X_ i \to Y'$ of $\mathcal{C}$ such that each $g_ i$ is the equivalence class of the pair $(f_ i : X_ i \to Y', s : Y \to Y')$.

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

$\xymatrix{ Y \ar[d]_{s_1} \ar[r]_{s_2} & Y_2 \ar[d]^{t_2} \\ Y_1 \ar[r]^{a_1} & Y' }$

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. $\square$

Comment #324 by arp on

Typo: In "Then $s = t_2 \circ s_1 \in S$ works" it should say $s = t_2 \circ s_2$.

There are also:

• 12 comment(s) on Section 4.27: Localization in categories

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).