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I am looking at Exercise 2.2.7 of Hatcher (pg 155):

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

Now what I don't understand in the question is the phrase "induced map on $H_n(\Bbb{R}^n,\Bbb{R}^n - \{0\})$. What does this mean? My interpretation is that $f: \Bbb{R}^n \to \Bbb{R}^n$ somehow gives a map $g$ from $H_n(\Bbb{R}^n,\Bbb{R}^n - \{0\})$ to itself.

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up vote 3 down vote accepted

Here is a restatement: $f$ induces a map of the pair $(R^n, R^n-{0})$ to itself. The $n$th homology of this pair is isomorphic to Z. The induced map on this homology group is $\pm 1$ depending on the sign of the determinant of $f$.

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You are quite right. You should read the section "Homotopy Invariance" starting on page 110. Hatcher begins by showing that continuous maps induce homomorphisms on (singular) chain groups. He then shows that these maps form a chain map from the singular chain complex of the source space to the singular chain complex of the target space. Finally, he shows that any chain map induces a homomorphism of homology groups.

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I know that my map $f$ will induce a map between $H_n(\Bbb{R}^n$ and itself. However now for the pair of spaces, $(\Bbb{R}^n,\Bbb{R}^n - \{0\})$ isn't the induced map ambiguous? –  user38268 Oct 6 '12 at 4:15
Hatcher defines induced maps on relative homology in the middle of page 118. –  Adam Saltz Oct 6 '12 at 13:29
In particular, $f$ induces a map from $\mathbb{R}^n \setminus {0}$ to itself because $f$ is invertible. –  Adam Saltz Oct 6 '12 at 13:32
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