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Proposition Let $f: A \hookrightarrow B$ and $f': A' \hookrightarrow B'$ be monomorphisms of simplicial sets. If $f$ is left anodyne, then the induced map

\[ \theta : (A \times B') \coprod _{ A \times A'} ( B \times A') \hookrightarrow B \times B' \]

is left anodyne. If $f$ is right anodyne, then $\theta $ is right anodyne.

Proof. We will prove the second assertion (the first follows by a similar argument). We proceed as in the proof of Proposition Let us first regard the monomorphism $f': A' \hookrightarrow B'$ as fixed, and let $T$ be the collection of all maps $f: A \rightarrow B$ for which the induced map

\[ \theta _{f,f'}: (A_{} \times B'_{} ) \coprod _{ A_{} \times A'_{} } ( B_{} \times A'_{} ) \hookrightarrow B_{} \times B'_{} \]

is right anodyne. We wish to show that every right anodyne morphism belongs to $T$. Since $T$ is weakly saturated, it will suffice to show that every horn inclusion $f: \Lambda ^{n}_{i} \hookrightarrow \Delta ^ n$ belongs to $T$ for $0 < i \leq n$. In this case, Lemma guarantees that $f$ is a retract of the morphism $g: (\Delta ^1 \times \Lambda ^ n_ i) \coprod _{ \{ 1\} \times \Lambda ^{n}_ i} ( \{ 1\} \times \Delta ^ n) \hookrightarrow \Delta ^1 \times \Delta ^ n$. It will therefore suffice to show that $g$ belongs to $T$. Replacing $f'$ by the monomorphism $(\Lambda ^{n}_{i} \times B'_{} ) \coprod _{ \Lambda ^{n}_{i} \times A'_{} } (\Delta ^ n \times A'_{} ) \hookrightarrow \Delta ^ n \times B'_{}$, we are reduced to showing that the inclusion $\{ 1\} \hookrightarrow \Delta ^1$ belongs to $T$.

Let $T'$ denote the collection of all morphisms of simplicial sets $f'': A''_{} \rightarrow B''_{}$ for which the map $(\{ 1\} \times B''_{} ) \coprod _{ \{ 1\} \times A''_{} } ( \Delta ^1 \times A''_{} ) \rightarrow \Delta ^1 \times B''_{}$ is right anodyne. We will complete the proof by showing that $T'$ contains all monomorphisms of simplicial sets. By virtue of Proposition, it will suffice to show that $T''$ contains the inclusion map $\operatorname{\partial \Delta }^ m \hookrightarrow \Delta ^{m}$, for each $m > 0$. In other words, we are reduced to showing that the inclusion $(\{ 1\} \times \Delta ^ m ) \coprod _{ \{ 1\} \times \operatorname{\partial \Delta }^ m } ( \Delta ^1 \times \operatorname{\partial \Delta }^ m) \hookrightarrow \Delta ^1 \times \Delta ^ m$ is right anodyne, which follows from Lemma $\square$