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Let $(X,x_0), (Y,y_0)$ be well-pointed spaces (inclusion of the basepoints is a cofibration). Show the following homotopy equivalence $$ \Sigma (X\times Y) \simeq \Sigma X \lor \Sigma Y \lor \Sigma (X\land Y), $$ where $\Sigma$ means a suspension (or reduced suspension if one prefers since it doesn't matter for well-pointed spaces) and $\land$ is a smash product.

Assuming we are using reduced suspension, it is quite clear that $\Sigma X \lor \Sigma Y$ is a subspace of the lhs, but I don't know how to somehow pull it outside and get $\Sigma (X\land Y)$.

Edit: I know the homotopy equivalence can be deduced from general theorems on "homotopy functors" (I hope that's how they are called in English) from chapter 7.7 in Spanier. But I was told there is an explicit proof and that's the one I'm looking for.

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As a possible method of attack, it might help to know that $\Sigma X\cong X\wedge S^1$ and for sufficiently well-behaved space (I think CW-complexes is enough), the smash product is commutative and associative up to homeomorphism. In particular, we can say that the RHS is homeomorphic to $\Sigma X\vee\Sigma Y\vee ((X\wedge Y)\wedge S^1)\cong\Sigma X\vee\Sigma Y\vee (X\wedge \Sigma Y)\cong\Sigma X\vee\Sigma Y\vee (\Sigma X\wedge Y)$. Whether this is a fruitful approach, I'm not sure. –  Daniel Rust Feb 12 '13 at 22:29
    
@DanielRust Associativity looks similar to $X\times(Y/\sim) \overset{?}{=} (X\times Y)/\sim$, so local compactness may (also) be sufficient here. Anyway, we can add general assumptions like local compactness or CW-stucture if it helps. –  savick01 Feb 13 '13 at 12:17
    
Does this result hold with $X$ and $Y$ are not path-connected? –  Colin Tan Nov 25 '13 at 6:31
    
@ColinTan I don't really know. You should go through details of the proof or ask Piotr. I tried to find a counterexample, but I failed. Trying to prove it, I ended up with a hypothesis $\sum({X\times Y}_{/X\times \{y_0\}}) = \sum Y \lor \sum(X\land Y)$. If it was true, we should be able to get a non-connected case by induction from the connected case (assuming finite number of connected components). It it was false and we had a counterexample, we probably could produce a counterexample for the original claim. –  savick01 Nov 25 '13 at 16:30

2 Answers 2

up vote 7 down vote accepted

I don't know how explicit we can go, but I'll give it a try. We have to go first through the homotopy-theoretical part.

Since $\{ * \} \subseteq X, \{ * \} \subseteq Y$ are cofibrations, $X \vee Y \subseteq X \times Y$ also is. Let $Z$ be a pointed space and consider the long exact sequence of homotopy for the pair $X \vee Y \subseteq X \times Y$, ie. the sequence

$\ldots \rightarrow [\Sigma ^{2}(X \vee Y), Z] \rightarrow [\Sigma(X \wedge Y), Z] \rightarrow [\Sigma(X \times Y), Z] \rightarrow [\Sigma(X \vee Y), Z] \rightarrow [X \wedge Y, Z] \rightarrow \ldots$,

where $[-,-]$ is the pointed set of homotopy classes of basepoint-preserving maps. Note that for any $n \geq 0$, $\Sigma^{n}(X \vee Y)$ is homeomorphic to $\Sigma^{n}X \vee \Sigma ^{n} Y$. I will not distinguish between the two.

Let $k \geq 1$ and define a map

$\psi ^{k}: \Sigma^{k}(X \times Y) \rightarrow \Sigma^{k}X \vee \Sigma^{k}Y$

$\psi ^{k} = \Sigma^{k}(i_{X} \pi_{X}) + \Sigma^{k}(i_{Y} \pi_{Y})$,

where $\pi: X \times Y \rightarrow X, Y$ are the projections and $i: X, Y \rightarrow X \vee Y$ are the inclusions. Addition is performed via the suspension structure on $\Sigma^{k}(X \times Y)$, so this is why we require $k \geq 1$. (Observe that even though I denote it by addition this is not necessarily commutative for $k=1$.)

If $j: X \vee Y \hookrightarrow X \times Y$ is the inclusion, then I claim that $\psi ^{k}$ is the left inverse to $\Sigma^{k}j$, ie. $\psi ^{k} \circ \Sigma^{k}j = id_{\Sigma^{k}(X \vee Y)}$. This is important because $\Sigma^{k}j$ are connecting maps in the long exact sequence of homotopy. Indeed, one computes

$\psi ^{k} \circ (\Sigma^{k}j) = (\Sigma^{k}(i_{X} \pi_{X}) + \Sigma^{k}(i_{Y} \pi_{Y})) \circ \Sigma^{k}j = \Sigma^{k}(i_{X} \pi_{X} j) + \Sigma^{k}(i_{Y} \pi _{Y} j) = \Sigma^{k}(id_{X} \vee const) + \Sigma^{k}(const \vee id_{Y}) \simeq (\Sigma^{k}id_{X} + const) \vee (const + \Sigma^{k}id_{Y}) \simeq \Sigma^{k}id_{X} \vee \Sigma^{k}id_{Y} \simeq id_{\Sigma^{k}X \vee \Sigma^{k}Y}$.

(One can also see this geometrically.) This immediately implies that for all $k \geq 1$ and all $Z$ the $[\Sigma^{k}(X \times Y), Z] \rightarrow [\Sigma^{k}(X \vee Y), Z]$ induced by $j$ is surjective and - by exactness of the long exact sequence - that for all $n \geq 1$ the map $[\Sigma^{n}(X \smash Y), Z] \rightarrow [\Sigma^{n}(X \times Y), Z]$ has zero kernel. In particular, for $k=1$ we have the short exact sequence of groups

$0 \rightarrow [\Sigma(X \wedge Y), Z] \rightarrow [\Sigma(X \times Y), Z] \rightarrow [\Sigma(X) \vee \Sigma(Y), Z] \rightarrow 0$

Moreover, the map induced by $\psi^{1}$ splits it and shows that there is a natural isomorphism

$\phi: [\Sigma(X \wedge Y), Z] \rtimes [\Sigma(X) \vee \Sigma(Y), Z] \rightarrow [\Sigma(X \times Y), Z]$,

of groups, where the product is only semi-direct, because our groups are not necessarily abelian. This is enough for our purposes, since we also have natural bijections

$[\Sigma(X \wedge Y), Z] \rtimes [\Sigma(X) \vee \Sigma(Y), Z] \simeq [\Sigma(X \wedge Y), Z] \times [\Sigma(X) \vee \Sigma(Y), Z] \simeq [\Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y), Z]$.

(The second one follows from from the fact that $\vee$ is the direct sum in the category of pointed spaces.) Yoneda lemma establishes that there is an isomorphism

$\theta: \Sigma(X \times Y) \rightarrow _{\simeq} \Sigma(X \smash Y) \vee \Sigma(X) \vee \Sigma(Y) $

in the homotopy category of pointed spaces, ie. a homotopy equivalence that we were after. It takes a little bookkeeping in the above Yoneda-lemma argumentation to see that such map is given by

$\theta = \Sigma(p) + \psi^{i} = \Sigma(p) + \Sigma^{1}(i_{X} \pi_{X}) + \Sigma^{1}(i_{Y} \pi_{Y})$,

where $p: X \times Y \rightarrow X \wedge Y$ is the natural projection. (This is what we get if we start with $id \in [\Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y), \Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y)]$ and trace it back by all the bijections above to $[\Sigma(X \times Y), \Sigma(X \wedge Y) \vee \Sigma(X) \vee \Sigma(Y)]$ - and this is the way to discover the isomorphisms "hidden" by Yoneda lemma.)

I understand that my exposition is far from perfect, but if you would like me to go into more detail over some parts, please comment.

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Great! Thank you very much. It seems that the problem is more complex than I expected / was told. I don't know the long exact sequence of a cofibration that you're using in the proof, can you give me any reference? It looks a bit similar to the exact sequence of homotopy groups of a fibration $Map(X\land Y,Z)\to Map(X\times Y,Z)\to Map(X\lor Y, Z)$ - can I deduce your sequence from this one? –  savick01 Mar 2 '13 at 17:39
    
It's very probable that there exist more elementary proofs, but this one is rather natural, since the basic property of all the spaces involved is that they form a cofibration sequence. One just proves that after one suspension this sequence "splits". The long exact sequence of homotopy for cofibrations I mention is purely dual to the one for fibrations. This is covered, for example, on page. 399 in Hatcher's "Algebraic Topology". Also, May in ch. 8 "Concise Course (...)" covers both of these long sequences on equal footing (highly recommended). –  Piotr Pstrągowski Mar 2 '13 at 18:30

This is proposition 4I.1 in Hatcher. The argument given there is geometric and elementary. It goes as follows:

We assume $X$ and $Y$ are CW complexes and we use the reduced suspension everywhere (which in this case is homotopy equivalent to the free suspension). I assume that your more general situation of well-pointed spaces works the same.

Consider the reduced Join $X*Y$ which is, like the regular join, the quotient of $X\times Y\times [0,1]$ by the relations

$$ (x,y_1,0)\sim(x,y_2,0) $$ $$ (x_1,y,1)\sim(x_2,y,1) $$

but we also collapse to a point the segment $(x_0,y_0,t)$. i.e. we collapse one "face" of the "cube" to $X$ and the opposite to $Y$ and the segment connecting the base points. Now, we glue in the cones on $X$ and $Y$ respectively along the collapsed faces and obtain a spcae

$$ Z=CX\sqcup_X (X*Y) \sqcup_Y CY $$

We will show that it is homotopy equivalent to the two spaces in your question.

On the one hand, if we collapse the two cones $CX$ and $CY$, what we get (by unwinding the definitions) is precisely the reduced suspension of $X\times Y$ (note that we used the reduced join). Since collapsing a contractible subcomplex does not change the homotopy type, we have a homotopy equivalence $Z\simeq \Sigma (X\times Y)$.

On the other hand, inside the reduced join $X*Y$, we have the spaces $X*y_0$ and $x_0*Y$ which are also contractible, as they are just cones on $X$ and $Y$ respectively, and collapsing them gives $\Sigma(X)\vee \Sigma(X\wedge Y)\vee \Sigma(Y)$. As before, this space is homotopy equivalent to $Z$ and we are done.

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