Hom functor is left-exact


In this post, we prove the well-used property that the Hom functor is left-exact. We consider the Hom functor of the following form $Hom_R(B,-)$ which is a functor from $R-mod$ the category of $R$-modules to $Ab$ the category of abelian groups. The functor contains the following data:


$A \mapsto Hom_R(B,A)$

$\{ f: A \rightarrow A' \} \mapsto \{ f^*: Hom_R(B,A) \rightarrow Hom_R(B,A') \}$

where $f^*$ is defined by $f^*(\phi) = f \circ \phi$. Then we see that Hom functor is a covariant functor.  
Proof) If $f : A \rightarrow A$ is the identity map, then $f^*(\phi) = f \circ \phi = f$, hence $f^*$ also becomes an identity map of $Hom_R(B,A)$. Also, let $f : A \rightarrow A'$ and $g: A' \rightarrow A''$, then we see that 
$(g \circ f)^*(\phi) = (g \circ f) \circ \phi  = g \circ (f \circ \phi) = g \circ (f^*(\phi)) = (g^* \circ f^*)(\phi)$. 
i.e. $Hom_R(-,B)$ is a covariant functor.

Now to our main point that $Hom_R(B,-)$ is left-exact. A functor $\mathcal{F}$ is left-exact if we have a exact sequence (of $R$-modules)

$0 \rightarrow A \stackrel{f}{\rightarrow} A' \stackrel{g}{\rightarrow} A'' \stackrel{h}{\rightarrow} 0$

then

$0 \rightarrow \mathcal{F}(A) \stackrel{f^*}{\rightarrow} \mathcal{F}(A') \stackrel{g^*}{\rightarrow} \mathcal{F}(A'')$

is exact.

1) $f^*$ is injective. Let $f^*(\phi) = f^*(\psi) \Rightarrow f \circ \phi = f \circ \psi$. If $f(\phi(a)) = f(\psi(a))$, then $f$ injective implies that $\phi(b) = \psi(b)$ for all $b \in B$, hence $\psi = \phi$. i.e. $f^*$ is injective.

2) 

($\subset$): $ker(g^*) = im(f^*)$. Let $\phi \in ker(g^*)$, then $g \circ \phi = 0$, then for all $b$, $\phi(b) \in ker(g) = im(f)$. There exists $c_b$ such that $\phi(b) = f(c_b)$. Define $\psi : B \rightarrow A$ by $b \mapsto c_b$, then $\phi(b) = f(c_b) = (f \circ \psi)(b)$ for all $b$, hence $\phi  = f \circ \psi \in im(f^*)$. 

($\supset$): Conversely, if $\phi \in im(f^*)$, then $\phi = f \circ \psi$ for some $\psi : B \rightarrow A$. Then for all $b \in B$, $\phi(b) = f(\psi(b)) \in im(f) = ker(g)$. This implie that $g \circ \phi(b) = 0$ for all $b$, i.e. $g \circ \phi = 0 \Rightarrow \phi \in ker(g^*)$.

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