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Let $X$ and $Y$ be nice spaces (connected, path-connected, locally contractible, etc). We do not assume they are CW complexes.

There is a natural inclusion map $$X\vee Y \rightarrow X\times Y.$$ Points in $X$ get mapped to $X\times \{y_0\}$, and points in $Y$ get mapped to $\{x_0\}\times Y.$

We know from the Kunneth theorem that $H_1(X\times Y)\cong H_1(X)\oplus H_1(Y).$ We also know from Mayer-Vietoris that $H_1(X\vee Y)=H_1(X)\oplus H_1(Y)$.

How do we see that this inclusion map induces the "obvious" isomorphism on the homology groups that sends $(\alpha,\beta)\rightarrow (\alpha, \beta)$?

This is perhaps a silly question, but I'm uncomfortable with the mechanics of the Kunneth formula (I use it as a black box, basically), so I'm having trouble making this "obvious" fact precise.

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    $\begingroup$ I think you mean $X \vee Y$ instead of $X \wedge Y$ here. $\endgroup$ – Rolf Hoyer Aug 10 '15 at 1:26
  • $\begingroup$ @RolfHoyer Indeed! Thank you for spotting that. $\endgroup$ – Potato Aug 10 '15 at 1:28
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    $\begingroup$ If only \wedge gave us the wedge in wedge product... $\endgroup$ – Ben West Aug 10 '15 at 1:56
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    $\begingroup$ How did you prove the Künneth isomorphism? It is more or less impossible to answer your question without knowing what map you have in mind. $\endgroup$ – Mariano Suárez-Álvarez Aug 10 '15 at 2:01
  • $\begingroup$ @MarianoSuárez-Alvarez My source for this material is Hatcher. He proves the Kunneth isomorphism for CW complexes using the cellular definition of the cross product, then uses CW approximation to prove it for general spaces. $\endgroup$ – Potato Aug 10 '15 at 2:16
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It helps to recall the formulation of the Kunneth theorem more fully.

While the fact that $H_1(X \times Y) = H_1(X) \oplus H_1(Y)$ is true here, this is a particular fact for dimension $1$. The Kunneth isomorphism shows that $H^*(X \times Y) \simeq H^*(X) \otimes H^*(Y)$, with the isomorphism being one of graded rings. This means that $H^k(X \times Y) = \bigoplus_{i=0^k} H^i(X) \otimes H^{k-i}(Y)$. In dimension $1$, this works out to $H^1(X \times Y) = (H^1(X) \otimes H^0(Y))\oplus (H^0(X) \otimes H^1(Y))$, and because we are assuming the spaces to be connected, we have that $H^0(X) \simeq H^0(Y) \simeq \mathbb{Z}$, and so this reduces to the isomorphism you stated above.

But the presence of the tensor product is important to the geometric understanding. It says (moving to homology from cohomology) that to build a $k$-cell in $H_k(X \times Y)$, you may take the product of an $i$-cell in $X$ and a $k -i$-cell in $Y$. So to create a loop in $X \times Y$, you can take a loop in one factor and a point in the other. This is like taking $\gamma \subseteq X$ and mapping it to $(\gamma \times \{y_0\}) \subseteq Y$ or vice versa. Note that this is the image of $\gamma$ under your map. In order to show that this induces an isomorphism on the homology (in particular, to see that every loop can be moved to be constant on one factor), it's necessary to appeal to the Kunneth theorem.

NB: the Kunneth theorem has all sorts of caveats. It's not true for all spaces, even reasonably nice ones. There are several formulations, but you usually have to assume there's no torsion in the homology groups considered or something similar.

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