# Exactness of Mayer Vietoris Sequence I am confused in showing that the sequence is exact at $$A_{n-1}' \oplus B_{n-1}$$. Here is part of my argument.

Note that $$(\rho_4+f_5)\circ (f_4,-\delta_4)=\rho_4 \circ f_4-f_5 \circ \delta_4=0$$ by commutativity of the diagram. This shows that $$\text{im}(f_4,-\delta_4) \subset \text{ker}(\rho_4+f_5)$$. Conversely, let $$(a',b) \in \text{ker}(\rho_4+f_5)$$. Then $$\rho_4(a')+f_5(b)=0 \implies f_5(b)=-\rho_4(a')$$.

I have no idea how to proceed. Can someone give me some hint? Thanks!

It's too awkward to keep track of the subscripts on the morphisms, so I'll write all $$\delta_j$$ as $$\delta$$ etc. I also assume that each $$f:C_k\to C'_k$$ is an isomorphism, so in particular, $$f:C_{n-1}\to C'_{n-1}$$ is an isomorphism.
You have $$f(b)=-\rho(a')$$. Then $$0=-\rho(\rho(a'))=-\rho(f(a'))=f(\delta(b)).$$ As $$f:C_{n-1}\to C'_{n-1}$$ is an isomorphism, $$\delta(b)=0$$ and by exactness, $$b=-\delta(a^*)$$ where $$a^*\in A_{n-1}$$. Then $$\rho(f(a^*)-a')=\rho(f(a^*))-\rho(a')=f(\delta(a^*)+b)=0$$ so $$a^*-a'=\rho(c')$$ by exactness. Then $$c'=f(c^*)$$ since $$f:C_n\to C_n'$$ is an isomorphism, and $$a^*-a'=f(\delta(c))$$. I would try $$-\delta(c)+a^*$$ as a candidate for an element of $$A_{n-1}$$ mapped to $$(a',b)$$.