Is wedge sum for finite CW complexes cancellative in the homotopy category?

Question

Let $X,Y,Z$ be finite pointed CW complexes. Is it possible that $X\vee Z$ and $Y\vee Z$ are homotopy equivalent, but $X$ and $Y$ are not?

Remark 1: Without the finiteness assumption on $Z$, there are silly examples like $X=pt$ , $Y$ some non-contractible complex and $Z$ a wedge of infinitely many copies of $Y$. If we only assume $Z$ is finite, this is still not trivial to me, but less interesting.

Remark 2: (Edit: there is a mistake in quotation here. We actually need $G$ and $H$ to be finite). By Van Kampen, we have $\pi_1(X\vee Z)\simeq \pi_1(X)*\pi_1(Z)$ where $*$ denotes the free sum of groups. As proved (very elegantly) here, for finitely generated groups it is true that $G*K\simeq H*K$ implies $G\simeq H$. Hence, one can't use $\pi_1$ to distinguish between $X$ and $Y$ in such a case.

Remark 3: One can exclude silly examples by imposing other finitness conditions on $X,Y$ and $Z$. For example, that they are $\pi$-finite (have finitely many non-trivial homotopy groups all of which are finite).

Having said that, I am mainly interested in the case where $Z=S^2$, but all kinds of examples and/or observations about the problem are welcome.

Answer 1

It is not a solution, but an observation in the case where $Z=S^2$.

Assume that $X,Y$ are CW complexes and that $X \vee S^2 \simeq Y \vee S^2 $. So there is a homotopy equivalence $f \colon X \vee S^2 \to Y \vee S^2$. This map induces a map $\hat{f} \colon X\to Y $ which is given by $p \circ f \circ i$ where $i$ is the inclusion of $X\subset X\vee S^2$ and $p$ is the map collapsing $S^2\subset Y\vee S^2$.

Further $f$ gives an isomorphism $$ \tilde{H}_*(X ) \oplus \tilde{H}_*(S^2) \cong \tilde{H}_*(X \vee S^2) \cong \tilde{H}_*(Y \vee S^2) \cong \tilde{H}(Y) \oplus \tilde{H}_*(S^2) .$$

This shows that, in degrees not equal to $2$, the induced map $\hat{f}$ is an isomorphism.

If in addition $H_2(X)$ is finite, then $\hat{f}$ is a homology isomorphism.

Now if we assume that $X$ and $Y$ are simple, then the dual whitehead theorems tell us that $\hat{f}$ is a homotopy equivalence.

Answer 2

Yes, it is possible, and so wedge sum for finite CW complexes does not satisfy homotopy cancellation. There is a description of some examples in the article "On principal $S^3$-bundles over spheres" by P. Hilton and J. Roitberg, although they mention they already appear in "On the Grothendieck group of compact polyhedra" by P. Hilton. Here is the idea of the construction:

Given a pointed map $\alpha \colon S^{m-1} \to S^n$ consider the mapping cone $C_{\alpha} = S^n \cup_{\alpha} D^m$. Assume that $\alpha$ is a suspension and the class $[\alpha] \in\pi_{m-1}(S^n)$ has prime order $p \neq 2,3$. Let $l$ be prime to $p$ and not congruent to $\pm 1$ mod $p$. Then, if $\beta \colon S^{m-1} \to S^n$ is a pointed map representing the class $ l[\alpha]$, then

$$ C_{\alpha} \vee S^m \simeq C_{\beta} \vee S^m $$

$$ C_{\alpha} \vee S^n \simeq C_{\beta} \vee S^n$$

but $C_{\alpha}$ is not homotopy equivalent to $C_{\beta}$.

We can not use this construction for $S^2$, for there are no elements in $\pi_k(S^2)$ nor in $\pi_1(S^n)$ that are suspensions and have prime order.

On the other hand, the article "On the Grothendieck group of compact polyhedra" also contains the following result. If $X$, $Y$ and $A$ are compact connected polyhedra that are suspensions and such that $X \vee A \simeq Y \vee A$, then the homotopy groups of $X$ and $Y$ are isomorphic.

The dual question is also approached, whether Cartesian product satisfies homotopy cancellation and the answer is also no, examples are constructed in the article "On principal $S^3$-bundles over spheres" using a construction which is a dual in some sense to the one given above.

There are many articles which refer to these two and which contain similar results.