PlanetMath: example of exact functor

Proposition 1 $\hom(F,-)$ is an exact functor for any free module $F$ over a ring $R$ .

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 $\im(f)\subseteq \im(\alpha)$ , 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 $F\ne 0$ is free, $F$ has a basis, say $X$ . For each $x\in X$ , let $M_x:=\alpha^{-1}(f(x))$ . Each $M_x$ is non-empty. By the axiom of choice (one of its equivalents if necessary), there is a function $g$ from $X$ to $\bigcup \lbrace M_x\mid x\in F\rbrace \subseteq M$ 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 $F$ . $\qedsymbol$

Proof. [Proof of Proposition 1] Let

$\displaystyle \xymatrix{ 0 \ar[r] & A \ar[r]^{\alpha} & B \ar[r]^{\beta} & C \ar[r] & 0 }$

be a short exact sequence of $R$ -modules. We want to show that

$\displaystyle \xymatrix{ 0 \ar[r] & \hom(F,A) \ar[r]^{\alpha^*} & \hom(F,B) \ar[r]^{\beta^*} & \hom(F,C) \ar[r] & 0 }$

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 $f=0$ .
$\im(\beta^*)=\hom(F,C)$ :
By Lemma 1 above, for every $f:F\to C$ , there is a $g:F\to B$ such that $\beta \circ g = f$ , or $\beta^*(g)=f$ .
$\im(\alpha^*)=\ker(\beta^*)$ :
If $f\in \im(\alpha^*)$ , 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 $\im(\alpha^*)\subseteq \ker(\beta^*)$ .

On the other hand, pick $f\in \ker(\beta^*)$ , so $f(F)\subseteq \ker(\beta)=\im(\alpha)$ . Therefore, by Lemma 1 above, there is a $g:F\to A$ such that $\alpha\circ g=f$ . This means that $f\in \im(\alpha^*)$ .

Thus, $\hom(F,-)$ is exact. $\qedsymbol$

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