This a fairly basic question relating to Theorem 8.11 from Rotman's Algebraic topology text (See pp. 190-192). It's a narrow question about how to determine if a given mapping is part of an exact sequence.
Theorem
Proof Excerpt
Rotman proves (and I follow his reasoning) that the following diagram commutes and the vertical lines are isomorphisms:
Question
So we know that $$ f_* = \psi k_* \phi $$
where $\psi$ and $\phi$ are isomorphisms. It seems obvious that $f_*$ is the map in the exact sequence of the theorem for $p > 0$, but I'm not sure how to prove this. It would suffice to show that $f_*$ preserves exactness of the sequence. I know that the image and kernel of $f_*$ are isomorphic to that of $k_*$.
Can someone provide a very explicit explanation for why this is the case?
So you've got a commutative diagram as follows: The top row is the Mayer-Vietoris LES, and each of the vertical maps is an isomorphism (I've taken the liberty of flipping the direction of the isomorphisms). We want to show that the bottom row is exact as well.
To see this, it suffices to check the following: $$\text{im}(\partial'_{p+1}) = \ker(f_{*})$$ and $$\text{im}(f_{*}) = \ker(i_{*}).$$
Since $\varphi$ is an isomorphism, we have $x \in \text{im}(\partial'_{p+1})$ if and only if we have $x \in \text{im}(\partial'_{p+1} \circ \varphi) = \text{im}(\psi \circ \partial_{p+1}),$ where equality follows from the commutativity of the diagram.
Then, since $\psi$ is an isomorphism, we have $x \in \text{im}(\psi \circ \partial_{p+1})$ if and only if $\psi^{-1}(x) \in \text{im}(\partial_{p+1}) = \ker(k_{*}),$ where the equality follows from the exactness of the top row.
Now, since $\phi$ is an isomorphism, we have $\psi^{-1}(x) \in \ker(k_{*})$ if and only if $\psi^{-1}(x)$ is in $\ker(\phi \circ k_{*}) = \ker(f_{*} \circ \psi)$, where equality follows by commutativity of the central square. But since $\psi$ is an isomorphism, $\psi^{-1}(x)$ is in $\ker(f_{*} \circ \psi)$ if and only if $x$ is in $\ker(f_{*}),$ so we have the desired equality $\text{im}(\partial'_{p+1}) = \ker(f_{*})$.
The argument for $\text{im}(f_{*}) = \ker(i_{*})$ is very similar, so I leave it to you.