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example of exact functor
In this entry, we show the following proposition:
Proposition 1.
$\hom(F,)$ is an exact functor for any free module $F$ over a ring $R$.
Before proving this, let us prove something related to a free module:
Lemma 1.
If $f:F\to N$ is a module homomorphism where $F$ is free, then for any module homomorphism $\alpha:M\to N$ where $\operatorname{im}(f)\subseteq\operatorname{im}(\alpha)$, there is a module homomorphism $g:F\to M$ such that $\alpha\circ g=f$.
Proof.
The case when $F=0$ is trivial. Suppose now $F\neq 0$ is free, $F$ has a basis, say $X$. For each $x\in X$, let $M_{x}:=\alpha^{{1}}(f(x))$. Each $M_{x}$ is nonempty. By the axiom of choice (one of its equivalents if necessary), there is a function $g$ from $X$ to $\bigcup\{M_{x}\mid x\in F\}\subseteq M$ such that $g(x)\in M_{x}$. Since $X$ is a basis for $F$, this function can be extended to a module homomorphism from $F$ to $M$. By abuse of notation, let us use $g$ for the extension. Since $\alpha(g(x))=f(x)$ for all $x\in X$, this implies that $\alpha\circ g=f$ on all of $F$. ∎
Proof of Proposition 1.
Let
$\xymatrix{0\ar[r]&A\ar[r]^{{\alpha}}&B\ar[r]^{{\beta}}&C\ar[r]&0}$ 
be a short exact sequence of $R$modules. We want to show that
$\xymatrix{0\ar[r]&\hom(F,A)\ar[r]^{{\alpha^{*}}}&\hom(F,B)\ar[r]^{{\beta^{*}}}% &\hom(F,C)\ar[r]&0}$ 
is short exact. This amounts to establishing the following three equations:

$\ker(\alpha^{*})=0$:
If $\alpha^{*}(f)=0$, then $\alpha\circ f=0$, which implies that $f(x)=0$ for all $x\in F$. This means that $f=0$.

$\operatorname{im}(\beta^{*})=\hom(F,C)$:
By Lemma 1 above, for every $f:F\to C$, there is a $g:F\to B$ such that $\beta\circ g=f$, or $\beta^{*}(g)=f$.

$\operatorname{im}(\alpha^{*})=\ker(\beta^{*})$:
If $f\in\operatorname{im}(\alpha^{*})$, then there is $g:F\to A$ such that $\alpha\circ g=f$. Therefore, $\beta^{*}(f)=\beta\circ f=\beta\circ(\alpha\circ g)=(\beta\circ\alpha)\circ g=% 0\circ g=0$, showing that $\operatorname{im}(\alpha^{*})\subseteq\ker(\beta^{*})$.
On the other hand, pick $f\in\ker(\beta^{*})$, so $f(F)\subseteq\ker(\beta)=\operatorname{im}(\alpha)$. Therefore, by Lemma 1 above, there is a $g:F\to A$ such that $\alpha\circ g=f$. This means that $f\in\operatorname{im}(\alpha^{*})$.
Thus, $\hom(F,)$ is exact. ∎
Remark. The converse of this is not true. But we do have the following fact: $\hom(P,)$ is an exact functor iff $P$ is a projective module (over some ring $R$).
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