Lemma 1.3.4.3. Let $f: S \rightarrow T$ be a morphism of simplicial sets which is bijective on both vertices and edges. If both $S$ and $T$ satisfy condition $(\ast ')$ of Proposition 1.3.4.1, then $f$ is an isomorphism.
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Proof. We claim that, for every simplicial set $K$, composition with $f$ induces a bijection
\[ \theta _{K}: \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( K, S) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( K, T). \]
Writing $K$ as a union of its skeleta $\operatorname{sk}_ n(K)$, we can reduce to the case where $K$ has dimension $\leq n$, for some integer $n \geq -1$ (see Definition 1.1.3.1). We now proceed by induction on $n$. The case $n = -1$ is trivial (since a simplicial set of dimension $\leq -1$ is empty). Let us therefore assume that $n \geq 0$, so that Proposition 1.1.4.12 supplies a pushout diagram of simplicial sets
\[ \xymatrix@R =50pt@C=50pt{ \coprod \operatorname{\partial \Delta }^{n} \ar [r] \ar [d] & \coprod \Delta ^{n} \ar [d] \\ \operatorname{sk}_{n-1}(K) \ar [r] & K. } \]
It follows from our inductive hypothesis that the maps $\theta _{\operatorname{\partial \Delta }^{n}}$ and $\theta _{ \operatorname{sk}_{n-1}(K)}$ are bijective. Consequently, to show that $\theta _{ K }$ is bijective, it will suffice to show that $\theta _{ \Delta ^{n} }$ is bijective: that is, that $f$ induces a bijection on $n$-simplices. For $n \leq 1$, this follows from our hypothesis. To handle the case $n \geq 2$, we observe that there is a commutative diagram
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^{n}, S) \ar [d]^{ \theta _{ \Delta ^ n} } \ar [r] & \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^{n}_{1}, S ) \ar [d]^{ \theta _{ \Lambda ^{n}_{1} }} \\ \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^{n}, T) \ar [r] & \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Lambda ^{n}_{1}, T). } \]
Here the right vertical map is bijective by virtue of our inductive hypothesis, and the horizontal maps are bijective by virtue of our assumption that both $S$ and $T$ satisfy condition $(\ast ')$. It follows that the left vertical map is also bijective, as desired. $\square$