TITLE: Why do we use the smash product in the category of based topological spaces?
QUESTION [17 upvotes]: I was telling someone about the smash product and he asked whether it was the categorical product in the category of based spaces and I immediately said yes, but after a moment we realized that that wasn't right.  Rather, the categorical product of $(X,x_0)$ and $(Y,y_0)$ is just $(X\times Y,(x_0,y_0))$.  (It seems like in any concrete category $(\mathcal{C},U)$, if we have a product (does a concrete category always have products?) then it must be that $U(X\times Y)=U(X)\times U(Y)$.  But I couldn't prove it.  I should learn category theory.  Maybe functors commute with products or something.)  Anyways, here's what I'm wondering: is the main reason that we like the smash product just that it gives the right exponential law?  It's easy to see that the product $\times$ I gave above has $F(X\times Y,Z)\not\cong F(X,F(Y,Z))$ just by taking e.g. $X=Y=Z=S^0$.

REPLY [3 votes]: This way $\mathbf{hCW_*}$, the pointed homotopy category of CW complexes, becomes a closed symmetric monoidal category with tensor product $\wedge$ and unit $S^0$. This is important, because once we observe that the functor 
$$\Sigma^\infty: \mathbf{hCW_*}\to \mathbf{hSp}$$
to the stable homotopy category is lax monoidal, it would follow easily that $\Sigma^\infty$ sends monoids to monoids. In particular, since $S^0$ is obviously a commutative monoid, the sphere spectrum $\mathbb{S}$ is a commutative ring spectrum. In English, the sphere spectrum $\mathbb{S}$ defines a cohomology theory that has a skew-commutative cup product.