Theorem 4.5.5.8. Let $X$ and $Y$ be simplicial sets. Then the comparison map $c_{X,Y}: X \diamond Y \rightarrow X \star Y$ of Notation 4.5.5.3 is a categorical equivalence of simplicial sets.

**Proof of Theorem 4.5.5.8.**
Let $X$ and $Y$ be arbitrary simplicial sets; we wish to show that the collapse map $c_{X,Y}: X \diamond Y \rightarrow X \star Y$ is a categorical equivalence. By virtue of Lemma 4.5.5.10, we may assume without loss of generality that $X = \Delta ^1$. Note that the map the $c_{X,Y}$ fits into a commutative diagram of simplicial sets

Note that the morphism $c_{X,\Delta ^{0}} \times \operatorname{id}_{Y}$ is a categorical equivalence by virtue of Proposition 4.5.5.12 and Remark 4.5.2.7. Consequently, to show that $c_{X,Y}$ is a categorical equivalence, it will suffice to show that the square on the right is a categorical pushout diagram (Proposition 4.5.3.7). Note that the square on the left is a pushout diagram in which the horizontal maps are monomorphisms, hence also a categorical pushout diagram (Proposition 4.5.3.8). We are therefore reduced to showing that the outer rectangle is a categorical pushout square (Proposition 4.5.3.5).

Specializing now to the case $X = \Delta ^1$, we wish to show that the lower square in the commutative diagram

is a categorical pushout diagram. We first claim that the upper square is a categorical pushout: by virtue of Proposition 4.5.3.8, this is equivalent to the assertion that the induced map

is a categorical equivalence. This follows from Remark 4.5.2.7, since $\theta $ factors as a product of the identity map $\operatorname{id}_{Y}$ with the inner horn inclusion $\Lambda ^{2}_{1} \hookrightarrow \Delta ^2$. To complete the proof, it will suffice to show that the outer rectangle is a categorical pushout diagram. Using the criterion of Proposition 4.5.3.8, we are reduced to showing that the map

is a categorical equivalence. Unwinding the definitions, we can identify $\rho $ with the composition

Here the map $\rho '$ is a categorical equivalence by virtue of Proposition 4.5.5.12 (together with Remark 4.5.3.10), and the map $\rho ''$ is inner anodyne by virtue of Proposition 4.3.6.4. $\square$