# A basic question on relative singular groups\homology

I have some basic questions on relative singular homology. Let $A \subset X$ and let $C_k(X,A)$ be the relative singular k-chains. Then it is said that since $\partial$ carries both $C_k(X)$ and $C_k(A)$ to $C_{k-1}(X)$ and $C_{k-1}(A)$ then $\partial$ descends to a map on the relative chain group. I tried to understand how to explicitly write this map. I like to represent elements of this quotient group as $\sigma^k_{X} + \{\beta^k_{A}\}$ where $\sigma^k_{X}$ is in $C_k(X)$ and so on. Then does the induced map $\partial$ carry $\sigma^k_{X} + \{\beta^k_{A}\}$ to $\partial\sigma^k_{X} + \{\partial\beta^k_{A}\}$ or to $\partial\sigma^k_{X} + \{\beta^{k-1}_{A}\}$. Also is there any conceptual problem with thinking the elements of this relative chain as in the coset form I have written there or is it better to handle them as completely new objects. I will ask my remaining questions depending on the answer to this one so I will keep them for now.

Thanks alot

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I guess it doesn't really matter, I'd write the image of $\sigma^k_{X} + \{\beta^k_{A}\}$ as $\partial \sigma^k_{X} + \{\partial \beta^k_{A}\}$, in your notation. But I'd think it's also ok to call it $\sigma^k_{X} + \{\beta^{\prime k-1}_{A}\}$ for some $\beta^{\prime k-1} \in C_{k-1}(A)$.

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my problem with $\partial \sigma^k_{X} + \{\partial \beta^k_{A}\}$ is that, it is not really the coset $\partial \sigma^k_{X} + \{\beta^{k-1}_{A}\}$ since $\{\partial \beta^k_{A}\} \subset \{\beta^{k-1}_{A}\}$ and that confuses me. $\partial \sigma^k_{X} + \{\partial \beta^k_{A}\}$ is not even a coset in $C_{k-1}(X,A)$ –  Sina Sep 21 '12 at 11:24
Example $H_{1} \subset G_{1}$ and $H_{2} \subset G_{2}$ are two groups and their subgroups and $\phi : G_{1} \rightarrow G_{2}$ is any homomorphism taking $H_{1}$ into $H_{2}$ then we get a well defined homomorphism (assuming$H_{1}$ and $H_{2}$ are both normal; which is true in abelian case) of the groups $G_{1}/H_{1} \rightarrow G_{2}/H_{2}$. –  s.b Sep 21 '12 at 21:17

By definition $C_{k}(X)$ is a free abelian group and $C_{k}(A)$ is a subgroup. The homomorphism $\partial : C_{K}(X) \rightarrow C_{k-1}(X)$ takes the sun-group $C_{k}(A) \rightarrow C_{k-1}(A)$. So by standard group theory (there is not topology here) it descends to a map of quoitents $C_{k}(X, A) \rightarrow C_{k-1}(X,A)$. So $\partial$ takes $\sigma_{X}^{k} + \beta_{A}^{k} \rightarrow \sigma_{X}^{k-1} + \beta_{A}^{k-1}$ where $\beta_{A}^{k-1} = \partial \beta_{A}^{k}$ and same for $\sigma.$

A minor point: You need some assumptions on the set $A$ (like it is closed, or a deformation retract of a neighborhod etc) to enure that $C_{k}(X) = C_{k}(A) \oplus \text{Nice complement}$ ut this is a not a strong assumption at all. But it is not true for arbitrary subsets $A$.

In my opinion, initially if you think it terms of simplicial homology (rather than singular homology) everything is much more transparent.

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Dear s.b., there is absolutely no need for any condition on the subspace $A$: it is a completely arbitrary subspace of $X$. The relevant complex is given by $C_k(X,A)=C_k(X)/C_k(A)$ and there is no need for a decomposition $C_{k}(X) = C_{k}(A) \oplus \text{complement}$ . You only need some condition on $A$ (e.g. that $A$ be a retract of $X$) if you want an isomorphism $H_k(X)\cong H_k(A)\oplus H_k(X,A)$ to hold. –  Georges Elencwajg Sep 20 '12 at 20:53
Yes you are right the boundary map descends to the quotient nicely for any subspace $A$ but the associated homology theory will not be nice unless some extra conditions on $A$ are fulfilled. –  s.b Sep 21 '12 at 21:06