I am referring to Theorem 6.3 p. 143 from Lang's Algebra (but the question description will be as self-contained as possible).
Let $A$ be a commutative ring and $M$, $W$, $V$, $U$ be $A$-modules. Let the sequence $0 \longrightarrow W \stackrel{\lambda}{\longrightarrow} V \stackrel{\phi}{\longrightarrow} U \longrightarrow 0 $ be exact.
By Proposition 2.1 p. 122, the induced sequence $0\longrightarrow Hom_A(U,M) \stackrel{\phi'}\longrightarrow Hom_A(V,M) \stackrel{\lambda'}\longrightarrow Hom_A(W,M) $ is exact. If we let $M$ be $A$ as a module over itself, then we obtain $0 \longrightarrow Hom_A(U,A) \stackrel{\phi'}{\longrightarrow} Hom_A(V,A) \stackrel{\lambda'}{\longrightarrow} Hom_A(W,A)$ exact or in Lang's notation $0 \longrightarrow U^{\vee} \stackrel{\phi'}{\longrightarrow} V^{\vee} \stackrel{\lambda'}{\longrightarrow} W^{\vee},$ where $V^{\vee}$ is the dual module of $V$.
Now, this is where i have a problem: it is mentioned in the proof of theorem 6.3 that since $A$ is projective (because it is free), then we also have exactness from the right, i.e. $0 \longrightarrow U^{\vee} \stackrel{\phi'}{\longrightarrow} V^{\vee} \stackrel{\lambda'}{\longrightarrow} W^{\vee} \longrightarrow 0$ is exact. I don't see where this exactness from the right comes from since $A$ being projective means that the functor $Hom_A(A,\cdot)$ is exact, while to obtain dual spaces we use the functor $Hom_A(\cdot,A)$.
Any insights? Thank you:-)