Lemma 4.6.2.18. Let $f: X \rightarrow Y$ be a morphism of Kan complexes which is fully faithful and essentially surjective. Then $f$ is a homotopy equivalence.

**Proof.**
Since $f$ is essentially surjective, the underlying map of connected components $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$ is surjective. We claim that it is also injective. To prove this, suppose that $x$ and $x'$ are vertices of $X$ such that $f(x)$ and $f(x')$ belong to the same connected component of $Y$. Then the morphism space $\operatorname{Hom}_{Y}( f(x), f(x') )$ is nonempty. Since $f$ is fully faithful, it induces a homotopy equivalence $\operatorname{Hom}_{X}(x,x') \rightarrow \operatorname{Hom}_{Y}(f(x), f(x') )$. It follows that $\operatorname{Hom}_{X}(x,x')$ is nonempty, so that $x$ and $x'$ belong to the same connected component of $X$. This completes the proof that $\pi _0(f)$ is a bijection.

By virtue of Whitehead's theorem (Theorem 3.2.7.1), it will suffice to show that for every vertex $x \in X$ having image $y =f(x) \in Y$ and every integer $n \geq 0$, the induced map $\theta : \pi _{n+1}( X, x) \rightarrow \pi _{n+1}( Y, y)$ is an isomorphism. Using Example 4.6.1.12, we can identify $\theta $ with the natural map $\pi _{n}( \operatorname{Hom}_{X}(x,x), \operatorname{id}_{x} ) \rightarrow \pi _{n}( \operatorname{Hom}_{Y}(y,y), \operatorname{id}_{y} )$, which is bijective by virtue of our assumption that $f$ induces a homotopy equivalence $\operatorname{Hom}_{X}(x,x) \rightarrow \operatorname{Hom}_{Y}(y,y)$. $\square$