Lemma 14.34.3. In Situation 14.34.1 the maps
1_ V \star \epsilon : V \circ X \to V, \quad \text{and}\quad \epsilon \star 1_ U : X \circ U \to U
are homotopy equivalences.
Proof. As in the proof of Lemma 14.34.2 we set Y = U \circ V so that we are in the situation of Example 14.33.1.
Proof of the first homotopy equivalence. By Lemma 14.33.4 to construct a map h : V \to V \circ X right inverse to 1_ V \star \epsilon it suffices to construct a map h_0 : V \to V \circ Y = V \circ U \circ V such that 1_ V = (1_ V \star d) \circ h_0. Of course we take h_0 = \eta \star 1_ V and the equality holds by (14.34.1.1). To finish the proof we need to show the two maps
(1_ V \star \epsilon ) \circ h, 1_ V \star \text{id}_ X : V \circ X \longrightarrow V \circ X
are homotopic. This follows immediately from Lemma 14.33.5 (with G = G' = V and F = F' = \text{id}_\mathcal {S}).
The proof of the second homotopy equivalence. By Lemma 14.33.4 to construct a map h : U \to X \circ U right inverse to \epsilon \star 1_ U it suffices to construct a map h_0 : U \to Y \circ U = U \circ V \circ U such that 1_ U = (d \star 1_ U) \circ h_0. Of course we take h_0 = 1_ U \star \eta and the equality holds by (14.34.1.1). To finish the proof we need to show the two maps
(\epsilon \star 1_ U) \circ h, \text{id}_ X \star 1_ U : X \circ U \longrightarrow X \circ U
are homotopic. This follows immediately from Lemma 14.33.5 (with G = G' = \text{id}_\mathcal {A} and F = F' = U). \square
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