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Let $C$ be a simplicially enriched category, i.e., there are

  • a collection of objects $ob C$,
  • a simplicial set $map_C(X,Y)$ for $X,Y \in ob C$,
  • composition maps $map_C(Y,Z) \times map_C(X,Y) \to map (X,Z)_C$, and
  • identity maps $\ast \to map_C(X,X)$

subject to associativity and unitality conditions. $C$ has an underlying category $C_0$ with morphism sets $C_0 (X,Y) = map_C(X,Y)_0$.

Moreover, assume that $C$ is simplicially tensored in a compatible way, i.e., there is a bifunctor $\otimes: C \times sSet \to C$ together with a natural isomorphism $$ map_C(X \otimes K, Y) \cong map_{sSet} (K,map_C(X,Y)). $$

In particular, any $n$-morphism $f \in map_C (X,Y)_n$ corresponds to a morphism $X \otimes \Delta^n \to Y$ of $C_0$. Given $(f: X \otimes \Delta^n \to Y) \in map(X,Y)_n$ and $(g: Y \otimes \Delta^n \to Z) \in map(Y,Z)_n$, there is an element $g \diamond f \in map(X,Z)_n$, namely the composition $$ X \otimes \Delta^n \overset{X \otimes diag}{\to} X \otimes (\Delta^n \times \Delta^n) \cong (X \otimes \Delta^n) \otimes \Delta^n \overset{f \otimes \Delta^n}{\to} Y \otimes \Delta^n \overset{g}{\to} Z $$ which uses the composition of the underlying category $C_0$ only.

Question: Is $g \diamond f = g \circ f$, i.e., is the composition rule $\circ$ completely determined by the tensoring and the composition on $C_0$?

This seems to be true if $C$ is additionally cotensored over $sSet$. At least, Hirschhorn seems to use this fact in "Model Categories and Their Localizations", but I could not find an explanation in his book or anywhere else.

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