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In studying degree of maps and homology groups (in Hatcher starting on p.134), I am having trouble separating the algebra from the underlying "shapes."

As a simple example, suppose a function maps a 2-dimensional sphere S2 around itself 3 times, giving deg = 3. This means the homology group is multiplied by 3, giving the new homology group {...-6,-3,0,3,6,...}.

But, the underlying shape has not changed topologically.

But, homeomorphic shapes have the same homology groups - thus a contradiction.

What am I doing wrong?

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    There is a map $*3: \mathbb{Z}\to\mathbb{Z}$. The existence of such a map does not mean that $\mathbb{Z}$ and $\mathbb{Z}$ are not isomorphic groups. It just means there is a map between them which is not an isomorphism. There is also a different map (the identity), which _is_ an isomorphism.2017-01-02
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    This has basically been said already but I just want to point out explicitly the fact that you seem to be confused about: *the subgroup $3\Bbb{Z}$ of $\Bbb{Z}$ is isomorphic to $\Bbb{Z}$*! So there's no issue here.2017-01-02
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    Yes, thanks. Viewing 3Z as distinct from Z was rather dumb on my part. But now I am confused about what the concept of degree does for us in terms of computing homology groups. Or is degree used primarily for other purposes?2017-01-02

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It seems you're blurring the distinction between a function and the image of a function. The homology group of $\mathbb{S}^2$ doesn't change; it just so happens that the function $\mathbb{S}^2 \to \mathbb{S}^2$ you describe induces a map on homology groups which is not the identity; no contradictions there.

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    Pls see my comment above.2017-01-02
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    @user54301: Degree concerns maps rather than spaces themselves, although it is a useful tool for deducing properties of spaces and maps between spaces via their homology groups. This is typical of any algebraic invariant: knowing only the algebraic invariant doesn't tell you much, but in conjunction with other information you can prove a lot of stuff.2017-01-02