Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have two simplicial complexes A and B, and A is a subcomplex of B. We know that there is an inclusion map from A to B, and I understand how to get the simplicial homology groups of each individual complex, but I am interested in the image of induced homorphism between the homology groups. Can someone explain to me what this is and how to compute it? Thanks!

share|cite|improve this question
up vote 9 down vote accepted

An element $\alpha\in H_*(A)$ of a homology group of $A$ is an equivalence class of linear combinations of simplices of $A$ —pick a representative of that class and call it $a$. Now $a$ is a linear combination of simplices of $A$, and since $A$ is a subcomplex of $B$, we can just as well see $a$ as a linear combination of simplices of $B$. In fact, viewed in this way it is a cycle, and therefore it represents an element $\beta\in H_*(B)$. It is not difficult to see that this element depends only on $\alpha$ and not on the particular representative $a$ we picked, so in this way we get a function $H_*(A)\to H_*(B)$ which maps each $\alpha$ to the corresponding $\beta$. This is the induced map, and this description makes it clear how to compute it (in principle).

Now, given an actual simplicial complex $A$, it is very rare that we compute $H_*(A)$ by actually writing its elements as homology classes of cycles: we do all sort of elaborate tricks, starting from the various long exact sequences to the more elaborate spectral sequences to invoking all sort of weird structure, to avoid as much as possible doing that. So in an actual concrete situation it is rare that we know actual representatives of homology classes. If that is the case, then of course the above description of the induced map is not going to be of any help in computing it! Each such situation has to be handled in a special way, depending on how the homology groups were determined, what information exactly is available, and so on.

share|cite|improve this answer
Thanks for the help! – Andrew Feb 27 '12 at 3:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.