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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? :)

2 Answers2

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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.

Adam Saltz
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    What if the $X_i$'s are not path-connected? – Twnk Jan 12 '20 at 20:19
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    @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. – epsilon_delta Dec 26 '20 at 07:07
  • Also when considering $S^0$? Because that space is not path connected, seems like very annoying things can happen then. – Rich_Rich Dec 13 '21 at 12:23
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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 $$\pi_n(X_1\times X_2)\cong\pi_n(X_1)\oplus\pi_n(X_2).$$

Dan Rust
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