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 sitation 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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