Why is the smash product in a certain slice category associative? I am trying to show the following. Let $(V, \times , \ast)$ be a cartesian closed symmetric monoidal category (so the monoidal product is the ordinary product). Consider the category $V_\ast$ of based objets, objets under $\ast$, the terminal objet of $V$. One then defines the smash produt in $V_\ast$ of v and w to be the pushout
$$\require{AMScd} \begin{CD}
v \amalg w @>>> v \times w ;\\
@VVV @VVV \\
\ast @>>> v \wedge w.
\end{CD}$$
Appearently, this is supposed to follow because $V$ is cartesian closed and by this, $v \times -$ preserves colimits. But I don't see how this gives my proposition, unfortunately.
Anyhelp would be appreciated.
 A: I think the best way to proceed is to construct the internal hom for $\mathcal{V}_*$. Writing $[-, -]$ for the internal hom of $\mathcal{V}$, we have the following equaliser diagram in $\mathcal{V}$,
$$\mathrm{Hom}_* (V, W) \rightarrow [V, W] \rightrightarrows W$$
where one arrow $[V, W] \to W$ is induced by the distinguished point $1 \to V$, and the other arrow $[V, W] \to W$ is induced by the distinguished point $1 \to W$. The distinguished point of $\mathrm{Hom}_* (V, W)$ corresponds to the zero morphism $V \to W$, of course. One then has to establish the following natural isomorphism:
$$\mathrm{Hom}_* (U \wedge V, W) \cong \mathrm{Hom}_* (U, \mathrm{Hom}_* (V, W))$$
Thus,
\begin{align}
\mathrm{Hom}_* (U \wedge (V \wedge W), X)
& \cong \mathrm{Hom}_* (U, \mathrm{Hom}_* (V \wedge W, X)) \\
& \cong \mathrm{Hom}_* (U, \mathrm{Hom}_* (V, \mathrm{Hom}_* (W, X))) \\
& \cong \mathrm{Hom}_* (U \wedge V, \mathrm{Hom}_* (W, X)) \\
& \cong \mathrm{Hom}_* ((U \wedge V) \wedge W, X)
\end{align}
so we have a natural isomorphism $U \wedge (V \wedge W) \cong (U \wedge V) \wedge W$. One also has to check the various coherence axioms, but this should be straightforward.
It is also possible to bypass the construction of the internal hom and simply work with the natural bijection
$$\mathcal{V}_* (U \wedge V, X) \cong \{ \text{bipointed morphisms } U \times V \to X \}$$
where a bipointed morphism is a morphism in $\mathcal{V}$ that preserves the distinguished point in each variable separately. The aim is to show that both $U \wedge (V \wedge W)$ and $(U \wedge V) \wedge W$ represent the functor of tripointed morphisms.
I suppose it may be possible to verify the associativity of $\wedge$ using only the fact that $\times$ preserves colimits in each variable separately, but this looks considerably more tedious than either approach outlined above.
A: Zhen writes: "I suppose it may be possible to verify the associativity of ∧ using only the fact that × preserves colimits in each variable separately, but this looks considerably more tedious than either approach outlined above."
Actually it is straight forward: $U \wedge (V \wedge W)$ is the quotient of $U \times (V \wedge W)$ which identifies $* \times (V \wedge W)$ with $U \times * \times *$. But $U \times (V \wedge W)$ is the quotient of $U \times V \times W$ which identifies $U \times * \times W$ with $U \times V \times *$ (in particular $U \times * \times *$). It follows that $U \wedge (V \wedge W)$ is the quotient of $U \times V \times W$ which identifies $* \times V \times W$ with $U \times V \times *$ and $U \times * \times W$. The reasoning applies to $(U \wedge V) \wedge W$. In fact, this shows that both are isomorphic to the object $U \wedge V \wedge W$ which classifies tripointed morphisms.
By the way, this whole construction can be seen as a special case of a more general theory about monoidal monads (see for example arXiv/1205.0101) applied to the monoidal monad $X \mapsto * + X$.
Finally let me mention that we cannot apply this to $(\mathsf{Top},\times)$, since here $\times$ doesn't preserve colimits in each variable. In fact, the smash product of pointed spaces is not associative in general (proof?).
