Need clarification on Exercise 2.2.7 Hatcher
Solution 1
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 $f$.
Solution 2
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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Pedro
Updated on October 06, 2020Comments
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Pedro about 3 years
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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Qinxuan about 11 yearsI 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?
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Adam Saltz about 11 yearsHatcher defines induced maps on relative homology in the middle of page 118.
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Adam Saltz about 11 yearsIn particular, $f$ induces a map from $\mathbb{R}^n \setminus {0}$ to itself because $f$ is invertible.