Does the functor $\pi_n\colon \mathsf{Top}_* \to \mathsf{Grp}$ preserve products

algebraic-topologycategory-theoryhigher-homotopy-groupshomotopy-theory

One of the very first propositions about the fundamental group in Hatcher's book [Hat01] states that the fundamental group functor preserves finite products (it is not hard to see that the isomorphism provided in the proof is natural):

Proposition 1.12 [Hat01]. $\pi_1(X\times Y)$ is isomorphic to $\pi_1(X)\times\pi_1(Y)$ if $X$ and $Y$ are path-connected.

Is something like this true for the $n$-th homotopy functor $\pi_n$?

That this is indeed true is suggested in this answer to a different question: when does a functor map products into products?, but the statement is phrased as preserving products in $\mathsf{Set}$. If this really translates into preservation in $\mathsf{Grp}$, what is a good reference for this statement? If not, what additional assumptions can I make to make true? Is it true for compact spaces, simplicial complexes?

[Hat01] Allen Hatcher. Algebraic topology. Cambridge University Press, 2001.

Best Answer

I will just elaborate on Tyrone's comment.

For a source, Hatcher gives a very brief "proof" along these lines in AT as Proposition 4.2 on page 343. The only assumption he places on the spaces is that they are path-connected. He doesn't give many details because "in spirit" the argument is the same as his proof for $\pi_1$ except for the fact that we treat arbitrary products now instead of finite, but the details are mostly unchanged. My hunch is you will find a proof with more details in Spanier's "Algebraic Topology" if you have access to it, I currently don't have a copy on hand.


In this discussion all spaces and functions/homotopies are assumed to be pointed.

If $X = \prod_\lambda X_\lambda$ is an arbitrary product of topological spaces, then by the universal property a continuous function $Y \to X$ is equivalent to a set of continuous functions $\{ f_\lambda\colon Y \to X_\lambda\}_\lambda$. In fact there is a continuous bijection

$$ \Phi\colon \operatorname{Map}(Y, X) \cong \prod_\lambda \operatorname{Map}(Y, X_\lambda)$$

given explicitly by $\Phi(f)=\prod_\lambda (\rho_\lambda \circ f)$, where $\rho_\lambda$ is the projection onto the $\lambda$-th factor. As per Tyrone's comment to my answer $\Phi^{-1}$ will also be continuous if $Y$ is locally-compact, but we don't need it for this argument because in any case after taking homotopy classes $\Phi$ descends to a bijection $\overline{\Phi}\colon [Y, X]\cong \prod_\lambda [Y, X_\lambda]$ (there is something to prove here, consider how a homotopy can be defined coordinate-wise).

Now, in the case that $Y= S^n$ this says that $\pi_n(X) \cong \prod_\lambda \pi_n(X_\lambda)$ as sets. But note that the bijection is given by $$ \overline{\Phi}([f]) = \prod_\lambda[\rho_\lambda \circ f] = \prod_\lambda\pi_n(\rho_\lambda)([f]), $$ so by functoriality it is also a homomorphism. Explicitly

$$\begin{align}\overline{\Phi}([f]+[g]) &= \prod_\lambda \pi_n(\rho_\lambda)([f] + [g])\\ &= \prod_\lambda \big(\pi_n(\rho_\lambda)([f]) + \pi_n(\rho_\lambda)([g]) \big) \\ &= \prod_\lambda \pi_n(\rho_\lambda)([f]) + \prod_\lambda\pi_n(\rho_\lambda)([g])\\ &= \overline{\Phi}([f]) + \overline{\Phi}([g]). \end{align}$$

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