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.

Help please? :)

Solution 1:

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.

Solution 2:

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 $$\pi_n(X_1\times X_2)\cong\pi_n(X_1)\oplus\pi_n(X_2).$$