I'm having difficulty understanding the following. It appears as Exercise 7, p. 155 in Hatcher's Algebraic Topology: (this is not homework, by the way)
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.
I know $GL_n(\mathbb{R})$ has precisely two path components, the set of matrices with positive determinant, and those with negative determinant. Also, if two maps are homotopic through pairs, they induce the same map on homology.
So if $\det(f)>0$, it is path connected to $1$, and this path induces a homotopy from $f$ to $1$. Then we know the induced maps $f_*=1_*$, but $1_*$ is the identity by the functorial properties, so $f_*=1$ (on the homology group).
If $\det(f)<0$, then $f$ is homotopic to a linear transformation with $-1$ in the $(1,1)$ entry, $1$ along the main diagonal, and $0$ elsewhere. I know reflections of $S^n$ have degree $-1$, and this map kind of looks like a reflection, but why should this map induce $-1$ on the homology group?
Best Answer
By naturality, it suffices to prove the same statement for
$f_* : H_{n-1}(\mathbb{R}^n - \left\{0\right\}) \to H_{n-1} ( \mathbb{R}^n - \left\{0\right\})$ where $f$ can be assumed to be a reflection as you have noted.
Let $\overline{f}: S^{n-1} \to S^{n-1}$ denote the map which is restriction of $f$ to $S^{n-1}$. Let $i : S^{n-1} \to \mathbb{R}^n - \left\{0 \right\}$.
Then we have $f \circ i = i \circ \overline{f} $
Hence, $f_* \circ i_* = i_* \circ \overline{f}_*$ in homology.
Note that since $i_*$ is an isomoprhism (as $S^{n-1}$ is a deformation retraction of $\mathbb{R}^n - \left\{0\right\}$) and all the homology groups are isomorphic to $\mathbb{Z}$, $i_*$ must send the generator ($1$) to another generator i.e $(\pm 1$).
So, it is enough to prove the statement for the map $\overline{f}_* : H_{n-1}(S^{n-1}) \to H_{n-1}(S^{n-1}) $, where $\overline{f}$ is a reflection of $S^{n-1}$. Hence $\deg( \overline{f})$ is $-1$ and the claim follows.