According to prop. 2.9,
$$M'\xrightarrow u M \xrightarrow v M'' \rightarrow 0$$
is exact iff the dual sequence
$$0\rightarrow \operatorname{Hom} (M'',N)\xrightarrow{\bar{v}} \operatorname{Hom}(M,N)\xrightarrow{\bar{u}} \operatorname{Hom} (M',N)$$
is exact for all $A$-modules $N$.
I'm confused wrt the proof that exactness of the dual sequence at $\operatorname{Hom}(M, N)$ implies exactness at the corresponding link in the original sequence. It begins like this:
First of all, since $\bar{v}$ is injective for all $N$ it follows that $v$ is surjective. Next, we have $\bar{u} \circ \bar v = 0$, that is $v \circ u \circ f = 0$ for all $f: M'' \to N$.
I don't get where the claim $v \circ u \circ f = 0$ comes from—comparing domains and codomains it simply doesn't make sense. Considering that $\bar u$ and $\bar v$ act (contravariantly) by precomposition, I think what was meant was $f \circ v \circ u = 0$. Am I correct?