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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.