Again, Hom functor is left-exact

From the previous post, we have considered $Hom_R(B, -)$, but this time we want to consider $Hom_R(-,A)$. Instead of getting a covariant functor, we now get a contravariant functor (i.e. reverses the map).

Consider a map $f: P_1 \rightarrow P_2$, then the natural way to pass it to Hom-sets are as follows,

$f^*: Hom_R(P_2, A) \rightarrow Hom_R(P_1, A) (\dagger)$

$\phi \mapsto \phi \circ f$

If we try to construct the order way, 

$Hom_R(P_1, A) \rightarrow Hom_R(P_2, A)$

$\phi \mapsto \phi \circ f$?

$\phi \mapsto f \circ \phi$?

Both composition would not make sense, hence we take $(\dagger)$ as our $Hom_R(-,A)$ functor. And obviously $R$-module $A$, $P$ is sent to $Hom_R(P,A)$. We now prove that $Hom_R(-,A)$ functor is again a left-exact functor, which means the following,

Given a short exact sequence,

$0 \rightarrow P_1 \stackrel{f}{\rightarrow} P_2 \stackrel{g}{\rightarrow} P_3$

$0 \rightarrow Hom_R(P_3,A) \stackrel{g^*}{\rightarrow} Hom_R(P_2, A) \stackrel{f^*}{\rightarrow} Hom_R(P_1, A)$

is exact.

Even though, the proof is exactly the dual of the proof from the last post, 

1) $g^*$ is injective. This is obvious because if $\phi_1, \phi_2$ are two maps from $Hom_R(P_3, A)$, then $\phi_1 \circ g = \phi_2 \circ g \Rightarrow \phi_1 = \phi_2$ because $g$ is surjective, i.e. epimorphism.

2) We show that $im~g^* = ker~f^*$

($\subset$): Let $\phi \in im~g^*$, then there exists $\psi$ such that $\phi = \psi \circ g$, then because we know that $im~f \subset ker~g$, we know that $\psi \circ g \circ f = 0$, i.e. $\phi \circ f = 0$, hence $\phi \in ker~f^*$.

($\supset$): Let $\phi \in ker~f^*$. Consider $\psi \in Hom(P_3, A)$ constructed as follows, $\psi(a) = \phi(b_a)$ where $b_a$ is an element such that $g(b_a) = a$. Such exists because $g$ is surjective. Let $g(b_a) = a = g(b_a')$, then we have $g(b_a - b_a') = 0$, hence $b_a - b_a' \in ker~g = im~f$, hence there exists some $c$ such that $b_a - b_a' = f(c)$, then $\phi(b_a - b_a') = \phi(f(c)) = 0$ by the assumption above. Hence $\psi$ is well-defined. Then we see that $\phi = \psi \circ g$, hence inside $im~g^*$.