# Higher Homotopy Group of a Product of Spaces

I want to show that for two toplogical spaces $X_1,X_2$ and for $x_1\in X_1 , x_2 \in X_2$ we have an isomorphism between $\pi_n (X_1 \times X_2 , (x_1,x_2))$ and $\pi_n (X_1, x_1) \times \pi_n (X_2, x_2)$ for all $n$.

I saw something kind of like this in chapter 4 of Hatcher's book, but I'm not quite sure how to make it rigorous here.

This is proposition 4.2 in Hatcher (as long as all the $X_i$ are path-connected).

The product $X_1 \times X_2$ is defined so that a continuous function $Y \to X_1 \times X_2$ corresponds exactly to a pair of continuous functions $Y \to X_1$ and $Y \to X_2$. Therefore any map $f: S_n \to X_1 \times X_2$ corresponds to a unique pair $f_1:S_n \to X_1$ and $f_2: S_n \to X_2$. A homotopy of $f$ is a map $S^n \times I \to X_1 \times X_2$, so it splits uniquely into two homotopies of maps into $X_1$ and $X_2$. (You should also check that everything is kosher with the basepoints.)

So the short version is that this follows from the definition of the product.

• What if the $X_i$'s are not path-connected?
– Twnk
Commented Jan 12, 2020 at 20:19
• @Twink When you fix the base point of the space, you actually fix the path component. So actually, what happens on the other components does not matter or, better say, homotopy groups, cannot capture it. Commented Dec 26, 2020 at 7:07
• Also when considering $S^0$? Because that space is not path connected, seems like very annoying things can happen then. Commented Dec 13, 2021 at 12:23

The projection maps $p_i\colon X_1\times X_2\to X_i$ for $i=1,2$ are fiber bundles with canonical sections. This means that we get a long exact sequence in homotopy $$\cdots\to\pi_n(X_2)\to \pi_n(X_1\times X_2)\to \pi_n(X_1)\to\cdots$$ which, because of the section, reduces to split short exact sequences $$0\to\pi_n(X_2)\to \pi_n(X_1\times X_2)\to \pi_n(X_1)\to 0$$ and so as this sequence of abelian groups (for $n\geq 2$) is split, we get an isomorphism $$\pi_n(X_1\times X_2)\cong\pi_n(X_1)\oplus\pi_n(X_2).$$

• Doesn't require that the groups are free? Commented Aug 5, 2021 at 12:46
• It's split because you have the sections $\pi_n(X_1) \to \pi_n(X_1 \times X_2)$ induced from the canonical inclusion $X_1 \to X_1 \times X_2$. Commented Aug 5, 2021 at 14:26
• What's the long exact sequence in homotopy you're reffering to? Commented Aug 5, 2021 at 15:27
• Edit : Sorry, understood the sequence. Which are the sections? Commented Aug 5, 2021 at 17:37
• It's exactly what I wrote in the comment above. Check the definition of a short exact sequence being right-split, then you just apply the splitting lemma. Commented Aug 6, 2021 at 12:36