# Showing that $S^2\times S^2$ and $S^2\vee S^2\vee S^4$ are not homotopy equivalent.

Show that the spaces $X=S^2\times S^2$ and $Y=S^2\vee S^2\vee S^4$ are not homotopy equivalent.

I read somewhere that $\pi_1(X)$ and $\pi _1(Y)$ are trivial, and that $X$, $Y$ have isomorphic homology groups. However, the cohomology ring of $X$ is a nontrivial product, while the cohomology ring of $Y$ is trivial. Therefore, $X$ and $Y$ are not homotopy equivalent.

I'm trying to figure out the cohomology rings of $X,Y$. I found on Wikipedia that: $H^*(S^2 \times S^2;\mathbb{Z})= \mathbb{Z}$ for degree $0$, $\mathbb{Z}^2$ for degree $2$, and $0$ otherwise.

Also from Wikipedia I found that the cohomology of the wedge sum of connected spaces is the cohomology of the disjoint union of those spaces which is the direct sum of their cohomologies. So for $H^*(Y)$ I got: $H^*(Y) = H^*(S^2) \oplus H^*(S^2) \oplus H^*(S^4) = \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$ for degree $0$, $\mathbb{Z}\oplus\mathbb{Z}$ for degree $2$, $\mathbb{Z}$ for degree $4$, and $0$ otherwise.

From this, the cohomology rings of $X,Y$ are not isomorphic. Hence, $X$ and $Y$ are not homotopy equivalent.

Is this correct at all? Could someone explain to me a few things: (1) is the cohomology ring just a list of the cohomology of the group in different degrees? (2) is what I have here correct for the cohomology of $X$ and $Y$?

Let $\Lambda$ be a module over a principal ideal domain $R$. The Künneth Theorem states that

$$H^n(X\times Y; \Lambda) \cong \bigoplus_{p+q=n}H^p(X; \Lambda)\otimes H^q(Y; \Lambda) \oplus \bigoplus_{p+q=n+1}\operatorname{Tor}(H^p(X; \Lambda), H^q(Y; \Lambda)).$$

If $\Lambda = R$ and $R$ is a field, then $\operatorname{Tor}(H^p(X; R), H^q(Y; R)) = 0$ as $H^p(X; R)$ and $H^q(Y; R)$ are free $R$-modules (i.e. vector spaces over $R$). For $\Lambda = R = \mathbb{Z}$, there may be some contribution from the $\operatorname{Tor}$ terms.

In this case, $X = Y = S^2$ and

$$H^p(S^2; \mathbb{Z}) = \begin{cases} \mathbb{Z} & p = 0, 2\\ 0 & \text{otherwise}. \end{cases}$$

As $H^p(S^2; \mathbb{Z})$ is a free $\mathbb{Z}$-module for every $p$, we see that all the $\operatorname{Tor}$ terms vanish and therefore

$$H^n(S^2\times S^2; \mathbb{Z}) = \begin{cases} \mathbb{Z} & n = 0, 4\\ \mathbb{Z}\oplus\mathbb{Z} & n = 2\\ 0 & \text{otherwise}. \end{cases}$$

As for the wedge sum, what you stated is not correct, it only holds for reduced cohomology. That is, $\widetilde{H}^n(X\vee Y; \mathbb{Z}) \cong \widetilde{H}^n(X;\mathbb{Z})\oplus\widetilde{H}^n(Y;\mathbb{Z})$. So your calculation of the cohomology groups is correct, except in degree zero:

$$H^n(S^2\vee S^2\vee S^4; \mathbb{Z}) \cong \begin{cases} \mathbb{Z} & n = 0, 4\\ \mathbb{Z}\oplus\mathbb{Z} & n = 2\\ 0 & \text{otherwise}. \end{cases}$$

As $H^n(S^2\times S^2; \mathbb{Z}) \cong H^n(S^2\vee S^2\vee S^4; \mathbb{Z})$ for every $n$, your argument for showing that $S^2\times S^2$ and $S^2\vee S^2\vee S^4$ are not homotopy equivalent doesn't work.

So far we've only considered the cohomology groups of the two spaces, but one can also consider their cohomology rings.

Let $X$ be a topological space. For any principal ideal domain $R$, cup product endows $H^*(X; R) := \bigoplus_{n\geq 0}H^n(X; R)$ with the structure of a graded ring. If $f : X \to Y$ is a continuous map, then $f^* : H^*(Y; R) \to H^*(X; R)$ is a ring homomorphism, in particular, if $f = \operatorname{id}_X$, then $f^* = \operatorname{id}$. It follows that if $X$ and $Y$ are homotopy equivalent, then $X$ and $Y$ have isomorphic cohomology rings, not just cohomology groups.

If $\alpha, \beta$ denote the two generators of $H^2(S^2\times S^2; \mathbb{Z}) \cong \mathbb{Z}\oplus\mathbb{Z}$, then $\alpha\cup\beta \neq 0$. On the other hand, if $\gamma, \delta$ denote the two generators of $H^2(S^2\vee S^2\vee S^4; \mathbb{Z}) \cong \mathbb{Z}\oplus\mathbb{Z}$, then $\gamma\cup\delta = 0$. Therefore $S^2\times S^2$ and $S^2\vee S^2\vee S^4$ do not have isomorphic cohomology rings, so they are not homotopy equivalent.

• @PedroTamaroff: Thanks. – Michael Albanese May 27 '17 at 3:50
• No problem. ;) ${}$ – Pedro Tamaroff May 27 '17 at 5:11