understanding proof of excision theorem

Question

I am trying to understand proof of excision theorem in homology following Hatcher's algebraic topology. I came to the following simple question, which I was unable to understand completely.

For a topological space $X$, let $C_n(X)$ denotes the free abelian group on singular $n$-simplices $\sigma:\Delta^n\rightarrow X$ (continuous). Let $A,B$ be subspaces of $X$ such that their interiors cover $X$.

Q.1 Is it true that $C_n(A\cap B)=C_n(A)\cap C_n(B)$ when considered them as subgroups of $C_n(X)$?

According to Hatcher's book, let $C_n(A+B)$ denotes the sums of chains in $A$ and chains in $B$. The second simple question is

Q.2 Does it mean that $C_n(A+B)=C_n(A)+C_n(B)$ where $C_n(A),C_n(B)$ are considered as subgroups of $C_n(X)$?

I came to these questions, because I didn't understand the following statement in Hatcher:

The map $C_n(B)/C_n(A\cap B)\rightarrow C_n(A+B)/C_n(A)$ induced by inclusion is obviously an isomorphism because both quotient groups are free with basis the singular $n$-simplices in $B$ that do not lie in $A$.

(If the answer to both questionsi s yes, then quoted statement is just third isomorphism theorem; but I was wondering whether this justification is correct?)

Answer 1

Q.1 Is it true that $C_n(A\cap B)=C_n(A)\cap C_n(B)$ when considered them as subgroups of $C_n(X)$?

Yes. You can think of an element of $C_n(X)$ as a huge tuple of integers, one for each singular $n$-simplex of $X$ (and with all but finitely many of them being $0$). $C_n(A)$ is then just all such tuples which are $0$ on all singular $n$-simplices which are not contained in $A$. So $C_n(A)\cap C_n(B)$ is all such tuples that are $0$ on all singular simplices that are not contained in $A$ and also $0$ on all singular simplices that are not contained in $B$. That is, they can only be nonzero on singular simplices that are contained in $A\cap B$, so this is the same as $C_n(A\cap B)$.

Q.2 Does it mean that $C_n(A+B)=C_n(A)+C_n(B)$ where $C_n(A),C_n(B)$ are considered as subgroups of $C_n(X)$?

Yes, by definition.

As for the statement in Hatcher, you can understand it using the third isomorphism theorem, but it is worthwhile to understand Hatcher's justification. It is instructive to consider the following toy example. Consider $X=\{0,1,2\}$, with $A=\{0,1\}$ and $B=\{1,2\}$. In this case there are only three singular simplices in $X$, the three constant maps. Let's identify $C_n(X)$ with $\mathbb{Z}^3$, then, with the three coordinates being the coefficients of the three simplices. We then have $C_n(A)=\{(x,y,0):x\in\mathbb{Z}\}$, $C_n(B)=\{(0,y,z):y,z\in\mathbb{Z})$, $C_n(A\cap B)=\{(0,y,0):y\in\mathbb{Z}\}$, and $C_n(A+B)=C_n(X)=\{(x,y,z):x,y,z\in\mathbb{Z}\}$. We can now explicitly see that the quotient $C_n(B)/C_n(A\cap B)$ is isomorphic to $\mathbb{Z}$, by sending the coset of $(0,y,z)\in C_n(B)$ to $z$. Similarly, the quotient $C_n(A+B)/C_n(A)$ is isomorphic to $\mathbb{Z}$, by sending the coset of $(x,y,z)\in C_n(A+B)$ to $z$.

The general story is basically the same. Both $C_n(B)/C_n(A\cap B)$ and $C_n(A+B)/C_n(A)$ are isomorphic to the group you would get by only considering the coordinates of simplices that are in $B$ but not in $A$.