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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (1)
Comment #324 by arp on
There are also: