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Section 5.5 Products

Proof.
A map \(f \colon S^n \to \prod_\alpha X_\alpha\) is the same data as a collection of maps \(f_\alpha \colon S^n \to X_\alpha\text{.}\) The map
\begin{equation*} \pi_n\!\left(\prod_\alpha X_\alpha\right) \to \prod_\alpha \pi_n(X_\alpha), \qquad [f] \mapsto ([f_\alpha])_\alpha, \end{equation*}
is compatible with homotopies, hence well-defined, and is the desired isomorphism.

Remark 5.18.

Compare this with homology, where we have the KĂĽnneth theorem (over \(\mathbb{Z}\)):
\begin{equation*} H_n(X \times Y) \;\cong\; \bigoplus_{i = 0}^{n} H_i(X) \otimes H_{n-i}(Y) \;\oplus\; \bigoplus_{i = 0}^{n-1} \mathrm{Tor}\bigl(H_i(X), H_{n-i-1}(Y)\bigr). \end{equation*}

Example 5.19.

Is it possible that \(X \not\simeq Y\) but \(\pi_n(X) \cong \pi_n(Y)\) for all \(n\text{?}\) Yes. Take
\begin{equation*} X = S^2 \times \mathbb{R}P^3, \qquad Y = \mathbb{R}P^2 \times S^3. \end{equation*}
For \(n = 1\text{,}\) \(\pi_1(S^2 \times \mathbb{R}P^3) \cong \mathbb{Z}/2 \cong \pi_1(\mathbb{R}P^2 \times S^3)\text{.}\) For \(n \geq 2\text{,}\) using that \(S^3 \to \mathbb{R}P^3\) and \(S^2 \to \mathbb{R}P^2\) are coverings,
\begin{equation*} \pi_n(X) \cong \pi_n(S^2) \times \pi_n(\mathbb{R}P^3) \cong \pi_n(S^2) \times \pi_n(S^3) \cong \pi_n(\mathbb{R}P^2) \times \pi_n(S^3) \cong \pi_n(Y). \end{equation*}
However, \(X\) and \(Y\) are not homotopy equivalent, since they have different homology groups. For instance, using the KĂĽnneth theorem:
\begin{equation*} H_5(X) = H_5(S^2 \times \mathbb{R}P^3) \cong \bigoplus_{i+j=5} H_i(S^2) \otimes H_j(\mathbb{R}P^3) \;\oplus\; \bigoplus_{i+j=4} \operatorname{Tor}(H_i(S^2), H_j(\mathbb{R}P^3)). \end{equation*}
We have \(H_2(S^2)\cong\mathbb{Z}\text{,}\) \(H_3(\mathbb{R}P^3)\cong\mathbb{Z}\text{,}\) and all other homology groups vanish in the relevant degrees. The only non‑zero term in the first sum is \(i=2,j=3\text{,}\) giving \(\mathbb{Z}\otimes\mathbb{Z}\cong\mathbb{Z}\text{.}\) The Tor terms vanish because \(H_i(S^2)\) is free. Hence \(H_5(X)\cong\mathbb{Z}\text{.}\)
For \(Y = \mathbb{R}P^2 \times S^3\text{:}\)
\begin{equation*} H_5(Y) \cong \bigoplus_{i+j=5} H_i(\mathbb{R}P^2) \otimes H_j(S^3) \;\oplus\; \bigoplus_{i+j=4} \operatorname{Tor}(H_i(\mathbb{R}P^2), H_j(S^3)). \end{equation*}
Here \(H_*(\mathbb{R}P^2)\) is \(\mathbb{Z}, \mathbb{Z}/2, 0\) in degrees \(0, 1, 2\text{,}\) and \(H_*(S^3)\) is \(\mathbb{Z}\) in degrees \(0\) and \(3\text{.}\) For \(i+j=5\) the possibilities are \((i,j) = (2,3)\text{,}\) which vanishes because \(H_2(\mathbb{R}P^2) = 0\text{,}\) and \((0,5)\) and \((1,4)\text{,}\) which vanish because \(H_5(S^3) = H_4(S^3) = 0\text{.}\) For the Tor sum, \((i,j)=(1,3)\) gives \(\operatorname{Tor}(\mathbb{Z}/2,\mathbb{Z})=0\) and \((i,j)=(0,4)\) gives \(\operatorname{Tor}(\mathbb{Z},\mathbb{Z})=0\text{.}\) Thus \(H_5(Y)=0\text{.}\)
Therefore \(H_5(X)\not\cong H_5(Y)\text{,}\) so \(X\) and \(Y\) are not homotopy equivalent.
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