If one has a long exact sequence of $R$-modules $$ \cdots \to A \to B \xrightarrow{f} C \to D \to \cdots $$
and $f$ is known to be surjective, does this imply that $D\cong 0$ as $R$-modules? Or is this exact sequence is equivalent to one of the form $$ \cdots \to A \to B \xrightarrow{f} C \to 0 \to \cdots $$
I know the following is a theorem:
The sequence $B\xrightarrow{f} C \to 0$ is exact $\iff f$ is an epimorphism.
But I suppose I don't know if this similar but slightly different statement is true:
The sequence $B\xrightarrow{f} C \to D$ with $f$ an epimorphism is exact $\iff D \cong 0$.
(For context, I am using this in the Mayer-Vietoris sequence, where we've shown a map is surjective and would like to conclude that a certain later term must be zero.)
You've given a perfectly good counterexample in the comments. The sequence $$\cdots \xrightarrow{0} \Bbb{Z} \xrightarrow{\text{id}} \Bbb{Z} \xrightarrow{0} \Bbb{Z} \xrightarrow{\text{id}} \Bbb{Z} \to \cdots$$ is exact: the image of each map is the kernel of the next one. The map $\mathrm{id}:\mathbb{Z}\to\mathbb{Z}$ is surjective, but the object following it in the sequence is another $\mathbb{Z}$, not $0$.
In general, if you have a sequence $B\stackrel{f}\to C\to D$, exactness just says that the image of $f$ is equal to the kernel of the map $C\to D$. So, $f$ is surjective iff the map $C\to D$ is $0$ (so that its kernel is all of $C$). But this doesn't mean $D$ itself has to be $0$, since you can have a zero homomorphism to any module at all. If the sequence continues with a map $D\to E$, then exactness at $D$ would additionally tell you the map $D\to E$ must be injective (its kernel must be the image of the map $C\to D$, which we know is $0$).