Lemma 14.28.3. Let \mathcal{C} be a category. Suppose that U and V are two cosimplicial objects of \mathcal{C}. Let a, b : U \to V be morphisms of cosimplicial objects. Recall that U, V correspond to simplicial objects U', V' of \mathcal{C}^{opp}. Moreover a, b correspond to morphisms a', b' : V' \to U'. The following are equivalent
There exists a homotopy h = \{ h_{n, \alpha }\} from a to b as in Remark 14.28.2.
There exists a homotopy h = \{ h_{n, i}\} from a' to b' as in Remark 14.26.4.
Thus a is homotopic to b as in Remark 14.28.2 if and only if a' is homotopic to b' as in Remark 14.26.4.
Proof.
In case \mathcal{C} has finite products, then \mathcal{C}^{opp} has finite coproducts and we may use Definitions 14.28.1 and 14.26.1 instead of Remarks 14.28.2 and 14.26.4. In this case h : U \to \mathop{\mathrm{Hom}}\nolimits (\Delta [1], V) is the same as a morphism h' : \mathop{\mathrm{Hom}}\nolimits (\Delta [1], V)' \to U'. Since products and coproducts get switched too, it is immediate that (\mathop{\mathrm{Hom}}\nolimits (\Delta [1], V))' = V' \times \Delta [1]. Moreover, the “primed” version of the morphisms e_0, e_1 : \mathop{\mathrm{Hom}}\nolimits (\Delta [1], V) \to V are the morphisms e_0, e_1 : V' \to \Delta [1] \times V. Thus e_0 \circ h = a translates into h' \circ e_0 = a' and similarly e_1 \circ h = b translates into h' \circ e_1 = b'. This proves the lemma in this case.
In the general case, one needs to translate the relations given by (14.28.1.1) into the relations given in Lemma 14.26.2. We omit the details.
The final assertion is formal from the equivalence of (1) and (2).
\square
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