proof of excision theorem: commutativity of a diagram (Reference: Hatcher's algebraic topology, p. 124) Let $X=A\cup B$ where $A,B$ are open. Let $C_n(X)$ denotes the free abelian group on all singular $n$-simplices $\sigma:\Delta^n\rightarrow X$, and $C_n^{U}(X)$ denotes the subgroup of $C_n(X)$ generated by those $n$-simplices $\sigma:\Delta^n\rightarrow X$ whose image is in $A$ or $B$. 

(1) Inclusion $C_n^{U}(X)\rightarrow C_n(X)$ induces isomorphism on homology groups $H_n^U(X)\rightarrow H_n(X)$. 

This holds true even when we go modulo $C_n(A)$ in the inclusion $C_n^{U}(X)\rightarrow C_n(X)$, i.e. 

(2) Inclusion $C_n^{U}(X)/C_n(A)\rightarrow C_n(X)/C_n(A)$ induces isomorphism $H_n^U(X,A)\rightarrow H_n(X,A)$

Now $C_n^U(X)/C_n(A)=[C_n(A)+C_n(B)]/C_n(A)\cong C_n(B)/C_n(A\cap B).$ Consider first and third term in this, and their exact sequences:
$$ \cdots \rightarrow \frac{C_{n+1}^U(X)}{C_{n+1}(A)}\rightarrow^{\partial} \frac{C_{n}^U(X)}{C_{n}(A)}\rightarrow^{\partial} \frac{C_{n-1}^U(X)}{C_{n-1}(A)}\rightarrow \cdots$$
$$ \cdots \rightarrow \frac{C_{n+1}(B)}{C_{n+1}(A\cap B)}\rightarrow^{\partial} \frac{C_{n}(B)}{C_{n}(A\cap B)}\rightarrow^{\partial} \frac{C_{n-1}(B)}{C_{n-1}(A\cap B)}\rightarrow \cdots.$$
Now the $n$-th terms are isomorphic in both sequences for each $n$; my question is simple (may be stupid one, but I couldn't justify naturally): 
Q. why the diagram should be commutative if we draw vertical (upward) arrows of isomorphisms in two sequences.

Why I was trying for this question is that, only after this, I can say surely that the $n$-th homology groups of last two chains are isomorphic i.e. 
$$H_n^U(X,A)\cong H^n(B,A\cap B),$$
and then with (2) this completes proof of excision theorem.
 A: When in doubt for a question like this, just pick an element and see what happens!  Consider an element of $C_{n+1}(B)/C_{n+1}(A\cap B)$, which is the coset of some element $x\in C_{n+1}(B)$.  If we first apply the horizontal map $\partial$, we get the coset of $\partial x\in C_n(B)$.  If we then apply the vertical isomorphism, we get the coset of $\partial x$, considered as an element of $C^U_n(X)$ (which contains $C_n(B)$ as a subgroup).
On the other hand, if we first apply the vertical isomorphism to $x$, we get the coset of $x$ considered as an element of $C^U_{n+1}(X)$.  Then if we apply the horizontal map $\partial$, we get the coset of $\partial x$ as an element of $C^U_n(X)$.  Note that this is the same $\partial x$ as in the previous paragraph: in both cases, the map $\partial$ is just defined using the ordinary boundary of singular chains.
So, for any element of $C_{n+1}(B)/C_{n+1}(A\cap B)$, if we first go right and then up, that's the same as first going up and then right.  That means the square from $C_{n+1}(B)/C_{n+1}(A\cap B)$ to $C^U_n(X)/C_n(A)$ commutes.  Since this is true for every $n$, the entire diagram commutes.
