Although there are good articles about this theme like induced map homology example, I would like to get a more explicit answer.

I know that one way to find such a map is the following: $ f:X\to Y $, then $ f_\ast[x]=[f (x)] $. So we have to look at the generator of $ H_p(X) $ under $ f $ and express it in terms of generators of $ H_p (Y) $. I can easily imagine the generators of the case $ p=1$, since the homology group is then the abelization of the corresponding fundamental group, but for $ p\neq1$ I have no picture in mind. Could someone explain me how to find the generators of the other homology groups to calculate the induced map? How can I find the map in the other cases ($ p\neq1$)? An explicit example would be helpful.

An other way: We look at the CW-structure or at the simplicial structure and deduce the induced map on the homology.
I can't imagine how this way works. Could someone explain it by giving an explicit example?

This question often arises, when I use Mayer-Vietoris. Maybe an example in this context would be helpful. Thank you!

  • $\begingroup$ Do you understand how to compute the degree of a map $f : S^n \to S^n$? $\endgroup$ – Lee Mosher Aug 4 '15 at 13:36

The key example is $f : S^n \to S^n$ and you want to compute $$\mathbb{Z} \approx H_n(S^n;\mathbb{Z}) \xrightarrow{f_*} H_n(S^n;\mathbb{Z}) \approx \mathbb{Z} $$ This function is simply the homomorphism "multiply by $degree(f)$". And you can compute the degree very easily. For example:

  • If $f$ is an orientation preserving homeomorphism then $degree(f)=1$
  • If $f$ is an orientation reversing homeomorphism then $degree(f)=-1$
  • If $n=1$ and $f$ is the "$k$th power map" $f(z)=z^k$ then $degree(f)=k$.
  • By induction from lower dimensional examples such as the previous example, if $g : S^{n-1} \to S^{n-1}$ has degree $k$, and if $f : S^n \to S^n$ is the suspension of $g$ (equal to $g$ on the equator and on each latitudinal sphere (using appropriate spherical coordinates), and fixing the north and south poles), then $degree(f)=k$.
  • If $f$ is differentiable and $y$ is a regular value then $f^{-1}(y)$ can be written as $f^{-1}(y) = \{a_1,…,a_I\} \cup \{b_1,…,b_J\}$ where each $Df_{a_i}$ preserves orientation and each $Df_{b_j}$ reverses orientation, in which case $degree(f)=I-J$.

It turns out that this example is the key to all computation. These kinds of degree calculations extend first to relative homology of maps of pairs $(D^n,\partial S^{n-1}) \to (D^n,\partial S^{n-1})$, and then they extend to the CW chain complex of maps between CW complexes. And that's how CW chain maps are computed in practice.

To be specific about CW complexes, let $f : X \to Y$ be a continuous map between CW complexes. Before getting a good formula for the induced chain map, one must first one homotope $f$ so that $f$ takes the $d$-skeleton of $X$ to the $d$-skeleton of $Y$ for all $d$. Then, given a $d$-cell $e \subset X$ and a $d$-cell $e' \subset Y$, the matrix of the chain map $f_\# : C_d(X) \to C_d(Y)$ contains an $e,e'$ term. This term is computed as a degree. Roughly speaking, that term is the degree of the map of $e$ over $e'$. For instance, assuming that $f$ is smooth on $e \cap f^{-1}(e')$, then you can use the formula in the last bullet point, taking $y \in e'$ to be a regular value.

  • $\begingroup$ Thanks @LeeMosher! So if I have a CW-complex $ X$ then there is a cellular chain complex of $ X $: $...\to H_{n+1}(X^{n+1}, X^n)\overset{d_{n+1}}{\to} H_n (X^n, X^{n-1})\overset{d_n}{\to} H_{n-1}(X^{n-1}, X^{n-2})\to... $ where $ d_n =j_{n-1}\partial_n $ with $H_n (X^n, X^{n-1})\overset {\partial_n}{\to }H_{n-1}(X^{n-1})\overset {j_{n-1}}{\to} H_{n-1}(X^{n-1}, X^{n-2})$. Then $ d_n $ is given by the Cellular Boundary Formula (cf. math.stackexchange.com/questions/536570/…). Do you mean this? How does this gives me the induced map (for example in MV-sequence)? $\endgroup$ – Daniel Bernoulli Aug 4 '15 at 16:48
  • $\begingroup$ I was referring more to your original question, namely how does one compute induced homology maps. I added a final paragraph to clarify this. $\endgroup$ – Lee Mosher Aug 4 '15 at 17:00
  • $\begingroup$ But, on the other hand, one can use the same idea to compute the boundary map $d_n : H_n(X^n,X^{n-1}) \to H_{n-1}(X^{n-1},X^{n-2})$. Given an $n$-cell $e$ with attaching map $f : S^n \to X^{n-1}$, and given an $n-1$-cell $e'$, the $e,e'$ term of $d_n$ is the "degree of the $f$ over $e'$, obtained for example by choosing a regular value of $f$ over a point of $e'$. $\endgroup$ – Lee Mosher Aug 4 '15 at 17:03
  • $\begingroup$ OK. That enlightens me a bit. Could you give an example for the last paragraph of your answer by calculating the degree of a map exemplarily, @LeeMosher? I would be very pleased! $\endgroup$ – Daniel Bernoulli Aug 4 '15 at 17:25
  • $\begingroup$ Take a map of the torus $T^2=S^1 \times S^1$ by the formula $f : S^1 \times S^1 \to S^1 \times S^1$ given by $f(z,w)=(z^m,w^n)$. The 1-skeleton is $\{z=1\} \cup \{w=1\}$ and there is a unique 2-cell. The point $(z,w)=(-1,+1)$ is a regular value, it has $mn$ pre-images, and the derivative preserves orientation over each pre-image. So the induced map of $f$ on $H_1(T^2;\mathbb{Z}) \approx \mathbb{Z}$ is multiplication by $mn$. $\endgroup$ – Lee Mosher Aug 4 '15 at 18:14

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