How does the image of the Hurewicz map $\pi_n(X,x) \to H_n(X)$ depend upon the choice of the base point?

Question

Let $X$ be a path connected topological space. I understand that the homotopy groups $\pi_n(X,x_0)$ and $\pi_n(X,x_1)$ are isomorphic to each other. However I do not understand whether the image of the Hurewicz map $\pi_n(X,x) \to H_n(X)$ is dependent or independent of the choice of basepoint. Is there any easy way to understand this ? Apologies if I am asking something sily. I would greatly appreciate any references. Thanks.

Answer 1

By naturality, it suffices to examine the universal example $X = S^n$. In this case, the Hurewicz homomorphism $\pi_n(S^n, s) \to H_n(S^n)$ is an isomorphism for any base point $s$, so in particular the image does not depend on the base point.

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EDIT: Here is the naturality argument I had in mind. Let $f \in \pi_n(X, x)$ correspond to a map $f: (S^n, s) \to (X, x)$. Then we have $$\require{AMScd} \begin{CD} \pi_n(S^n, s) @>\cong>> H_n(S^n) \\ @V{f_*}VV @VV{f_*}V \\ \pi_n(X, x) @>>h> H_n(X) \end{CD} $$

We have $h(f) = f_*(1)$, for $f_*$ the induced map on homology which doesn't care about basepoints. This reduces the question to the issue of whether the classes $\{f: (S^n, s) \to (X, x_1)\}$ and $\{f: (S^n, s) \to (X, x_2)\}$ induce the same collection of maps in homology. Homology doesn't care about basepoints, and forgetting base points, we get the same classes of maps up to homotopy provided $x_1$ and $x_2$ are in the same path component of $X$, as $S^n$ is connected for $n > 0$.

Answer 2

Note that we don't just have some arbitrary isomorphism $\pi_n(X,x_1)\to \pi_n(X,x_0)$; we have an explicit description of what the map is. Namely, we can get such an isomorphism by picking a path $\gamma$ from $x_0$ to $x_1$ and then inserting copies of $\gamma$ radially starting at the basepoint $s_0$ of $S^n$ to turn a map $f:(S^n,s_0)\to (X,x_1)$ into a map $f^\gamma:(S^n,s_0)\to (X,x_0)$. Now the key observation is that this map $f^\gamma$ is actually homotopic to $f$ as a map $S^n\to X$ (i.e., ignoring the basepoints). The homotopy is messy to write down explicitly but easily to visualize: you just gradually shrink the radial extensions, using only the portion between $\gamma(t)$ and $x_1=\gamma(1)$ for the $t$th step of the homotopy (so the $t$th step maps $s_0$ to $\gamma(t)$). In terms of the picture at the top of page 341 of Hatcher's Algebraic Topology, the intermediate stages of the homotopy are given by restricting to squares which are intermediate between the inner $f$ square and the full outer square.

In particular, this means $f$ and $f^\gamma$ induce the same map on $H_n$. Since the image of $f$ under the Hurewicz map is just the image of the fundamental class in $H_n(S^n)$ under $f$, this means that $f$ and $f^\gamma$ have the same Hurewicz image. It follows that the Hurewicz images of $\pi_n(X,x_1)$ and $\pi_n(X,x_0)$ are the same.