Kerodon

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Proposition 7.2.1.5. Let $f: A \rightarrow B$ be a morphism of simplicial sets. Then:

$(1)$

If $f$ is either left cofinal or right cofinal, then it is a weak homotopy equivalence.

$(2)$

If $f$ is a weak homotopy equivalence and $B$ is a Kan complex, then $f$ is left and right cofinal.

Proof. We first prove $(1)$. Let $X$ be a Kan complex. Then the projection map $X \times B \rightarrow B$ is a Kan fibration (Remark 3.1.1.6), and therefore both a left and a right fibration (Example 4.2.1.5). Consequently, if $f$ is either left cofinal or right cofinal, the induced map

$\operatorname{Fun}(B,X) \simeq \operatorname{Fun}_{/B}(B, X \times B) \rightarrow \operatorname{Fun}_{/B}(A, X \times B) \simeq \operatorname{Fun}(A,X)$

is a homotopy equivalence of Kan complexes. Allowing $X$ to vary, we conclude that $f$ is a weak homotopy equivalence.

We now prove $(2)$. Assume that $B$ is a Kan complex and that $f$ is a weak homotopy equivalence; we will show that $f$ is left cofinal (the proof that $f$ is right cofinal is similar). Let $q: \widetilde{B} \rightarrow B$ be a left fibration. Since $B$ is a Kan complex, $q$ is a Kan fibration (Corollary 4.4.3.8); in particular, $\widetilde{B}$ is a Kan complex. Applying Corollary 3.1.3.4, we obtain a commutative diagram of Kan complexes

$\xymatrix@R =50pt@C=50pt{ \operatorname{Fun}(B, \widetilde{B} ) \ar [d]^{q \circ } \ar [r]^-{\circ f} & \operatorname{Fun}(A, \widetilde{B}) \ar [d]^{q \circ } \\ \operatorname{Fun}(B, B) \ar [r]^-{ \circ f} & \operatorname{Fun}(A,B ), }$

where the vertical maps are Kan fibrations (Corollary 3.1.3.2). Our assumption that $f$ is a weak homotopy equivalences guarantees that the horizontal maps are homotopy equivalences (Proposition 3.1.6.17). Applying Proposition 3.2.8.1, we deduce that the map $\operatorname{Fun}_{/B}( B, \widetilde{B} ) \rightarrow \operatorname{Fun}_{/B}(A, \widetilde{B})$ is also a homotopy equivalence. $\square$