# Smash product of CW complexes

Mathematics Asked by Lucas Giraldi A. Coimbra on December 23, 2020

I’m studying algebraic topology and I’m using Hatcher’s book. There, he talks about the smash product of CW-complexes:

Given two CW-complexes $$X$$ and $$Y$$ and two points $$x_0 in X$$ and $$y_0 in Y$$, the wedge product $$X vee Y$$ is the CW-complex in which $$x_0$$ and $$y_0$$ are identified, that is $$X vee Y = frac{X sqcup Y}{{x_0, y_0}}.$$ The smash product of $$X$$ and $$Y$$ is the CW-complex $$X wedge Y = frac{X times Y}{X vee Y}.$$

Is there a good way to see intuitively what’s happening in the smash product? Hatcher describes it as "collapsing away the parts that are not genuinely a product, the separate factors of $$X$$ and $$Y$$" but I’m struggling to understand this statement.

Let $$M,N$$ be locally compact Hausdorff spaces. Then their one-point compactifications $$M_infty,N_infty$$ are compact Hausdorff spaces, and each is equipped with a canonical basepoint. The product $$Mtimes N$$ is locally compact Hausdorff and we have the basic relation $$(Mtimes N)_inftycong M_inftywedge N_infty.$$

This is where the smash product comes from and how you should think about it intuitively. In many situations where one would like to do homotopy or algebraic topology, basepoints are not naturally available. The problem is getting into the pointed homotopy category so as to be able to fully exploit the power of many homotopy-theoretic constructions. For instance, think about using the formula above to understand the compactly supported cohomology of a product of open manifolds.

As an example, for each $$ngeq0$$, the sphere $$S^n$$ is the one-point compactification of $$mathbb{R}^n$$. In fact from the homotopy theorist's persepective this is a much better definition of the $$n$$-sphere since it gives it a canonical basepoint. As an equation the statement reads $$S^ncong(mathbb{R}^n)_infty$$, and from the above we immediately get for any $$m,ngeq0$$ that $$S^{m+n}cong (mathbb{R}^{m+n})_infty=(mathbb{R}^mtimesmathbb{R}^n)_inftycong S^mwedge S^n.$$

More special cases present themselves. If $$M$$ is compact Hausdorf, then $$M_infty=M_+=Msqcup{infty}$$ is a disjoint union of $$M$$ and an additional point. If $$N$$ is also compact Hausdorff then we see that the standard identity $$(Mtimes N)_+cong M_+wedge N_+$$ is just a very special case of the more general statement discussed above.

Note also that if $$X$$ is compact Hausdorf with basepoint $$x_0$$, then $$(Xsetminus{x_0})_inftycong X$$ as pointed spaces. Thus we have found a recipe to intuitively understand the smash product $$Xwedge Y$$ of any two pointed compact Hausdorff spaces $$X,Y$$.

Now, all this can be made functorial in a sense. If $$f:Mrightarrow N$$ is a map between locally compact Hausdoff spaces, then it induces a (not necessarily continuous) basepoint-preserving function $$f_infty:M_inftyrightarrow N_infty$$ in the obvious way. The pointed function $$f_infty$$ is continuous if and only if $$f$$ is a proper map. Since we are working with locally-compact Hausdorff spaces this is equivalent to $$f$$ being compact. That is, for each compact $$Ksubset N$$, the inverse image $$f^{-1}(K)$$ is compact in $$M$$.

So let $$mathcal{LH}$$ be the category of locally compact Hausdorff spaces and proper maps. Then the one-point compactification $$Mmapsto M_infty$$ is a functor $$mathcal{LH}xrightarrow{(-)_infty} Top_*$$ which happens to take values in the subcategory of pointed compact Hausdorff spaces.

The category $$mathcal{LH}$$ does not have categorical products. A categorical product would necessarily agree with the cartesian product, and since the projections $$Mxleftarrow{pr_M}Mtimes Nxrightarrow{pr_N}N$$ are not in general proper maps, the construction cannot be made in $$mathcal{LH}$$.

However the cartesian product does supply $$mathcal{LH}$$ with a monoidal product. This is to be likened to the smash product of pointed spaces. Indeed, the equation $$(Mtimes N)_inftycong M_inftywedge N_infty$$ is essentially the statement that the functor $$mathcal{LH}xrightarrow{(-)_infty} Top_*$$ is monoidal. If fact it is even nicer. The functor even preserves homotopy, of course with the correct notion of homotopy in $$mathcal{LH}$$ being proper homotopy.

Correct answer by Tyrone on December 23, 2020

The best way to understand the smash product is by its universal properties. One comes from its expression as a quotient. A map $$X times Y rightarrow Z$$ factors through $$X wedge Y$$, if and only if, $$X vee Y subset X times Y$$ is mapped to a single point. This is a useful criterion to create maps out of smash products.

Another useful universal property smash products have is that they satisfy something like a tensor-hom adjunction in the category of pointed spaces. We have $$operatorname{Map_*}(X wedge Y , Z) cong operatorname{Map}(X, operatorname{Map}(Y,Z))$$. This follows from the usual adjunction between product and hom in the unpointed category plus the universal property in the pragraph above, or explicitly $$((x,y) rightarrow f(x,y)) rightarrow (x rightarrow (y rightarrow f(x,y)))$$. So from this perspective, the smash product is just the thing that is adjoint to pointed mapping spaces. This is a very important perspective, for example it leads us to studying loop spaces, because maps from a suspension to $$Z$$ are the same as maps from the original space to $$Omega Z$$. From there, one is very close to discovering Puppe sequences, one of the most important results in elementary algebraic topology.

For the record, some people mistakenly say that smash product is the categorical product in the category of pointed spaces. This is wrong. In fact, the categorical product is still the normal product of spaces. Perhaps what causes this confusion for people is that in the category of sets, the categorical product is also the adjoint to hom, but this is not true in general as we've shown.

Answered by Connor Malin on December 23, 2020

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