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
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Comment #324 by arp on
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