How is naturality used here?

Question

I'm trying to understand why in order to prove the claim:

For an invertible linear transformation $f\colon\mathbb{R}^n\to\mathbb{R}^n$ show that the induced map on $H_n(\mathbb{R}^n,\mathbb{R}^n-\{0\})\approx \tilde{H}_n(\mathbb{R}^n-\{0\})\approx\mathbb{Z}$ is $1$ or $-1$ according to whether the determinant of $f$ is positive or negative.

it follows by naturality that it suffices to prove the statement for:

$$f_*: H_{n-1}(\mathbb{R}^n-\{0\})\to H_{n-1}(\mathbb{R}^n-\{0\})$$

I'm familiar with the commutative diagrams in Hatcher's Algebraic Topology on pag. 127-128 but I'm not sure how this one fits in to those diagrams. I'm assuming the following one is the one I should be using somehow (?)

Any advice would be much appreciated.

(the claim that naturality suffices is from here)

Answer 1

If the determinant of $f$ is positive, then $f$ is in fact homotopic to the identity, and what you want is easy. Suppose then that that determinant ios negative, so that $f$ is homotopic to a reflection. We can therefore suppose that $f$ is the map $(x_1,x_2,\dots,x_n)\mapsto (-x_1,x_2,\dots,x_n)$.

This $f$ restricts to the unit sphere $S^{n-1}$, and naturality gives us a commutative diagram

H_{n-1}(S^{n-1}) —> H_{n-1}(R^n-0)
   |                  |
   V                  V
H_{n-1}(S^{n-1}) —> H_{n-1}(R^n-0)

with horizontal arrows induced by the inclusion — and isomorphisms — and the vertical arrows induced by $f$. It follows from this that the map induced by $f$ on $H_{n-1}(R^n-0)$ is $-1$ exactly when then map it induces on $H_{n-1}(S^{n-1})$ is $-1$. Now do something with degrees to compute the map indauced on $H_{n-1}(S^{n-1})$.

That is one place where naturality helps you.

Next, you want to compute the map induce by $f$ on $H_n(R^n,R^n-0)$, and naturality gives us a commutative diagram

H_n(R^n) —> H_n(R^n,R^n-0) —>  H_{n-1}(R^n-0) —> H_{n-1}(R^n)
   |               |               |              |
   V               V               V              V
H_n(R^n) —> H_n(R^n,R^n-0) —>  H_{n-1}(R^n-0) —> H_{n-1}(R^n)

with rows long exact sequences for the pair $(R^n,R^n-0)$ and vertical arrows induced by $f$. The four vertices of this diagram are zeros (I am secretly using reduced homology), so the commutativity gets you the map you want.