# Contravariant Hom Functor is Left Exact

I am trying to prove the following statement:

Let $$A\stackrel{\alpha}{\longrightarrow}B\stackrel{\beta}{\longrightarrow} C \rightarrow0$$ be an exact sequence of $$R$$-module homomorphisms. Prove that the sequence $$0\rightarrow \text{Hom}_R(C,M)\stackrel{\beta^{*}}{\longrightarrow}\text{Hom}_R(B,M)\stackrel{\alpha^{*}}{\longrightarrow} \text{Hom}_R(A,M)$$ of $$\mathbb{Z}$$-module homomorphisms is exact.

This is my proof:

If $$A\stackrel{\alpha}{\longrightarrow}B\stackrel{\beta}{\longrightarrow} C \rightarrow0$$ is an exact sequence of $$R$$-module homomorphisms, then we know that $$\beta$$ is surjective and $$\beta \circ \alpha = 0$$.
In order to show that $$0\rightarrow \text{Hom}_R(C,M)\stackrel{\beta^{*}}{\longrightarrow}\text{Hom}_R(B,M)\stackrel{\alpha^{*}}{\longrightarrow} \text{Hom}_R(A,M)$$ is exact, we must show that $$\beta^{*}$$ is injective and $$\text{Ker}(\alpha^{*})=\text{Im}(\beta^{*})$$.

We claim that $$\text{Ker}(\beta^{*})$$ is trivial, or in other words, $$\beta^{*}$$ is injective.
We have the following:
\begin{align} \text{Ker}(\beta^{*}) &=\{\sigma \in \text{Hom}_R(C,M) \mid \beta^{*}(\sigma)=0\}\\ &=\{\sigma \in \text{Hom}_R(C,M) \mid \sigma \circ \beta =0\}\\ &=\{\sigma \in \text{Hom}_R(C,M) \mid (\sigma \circ \beta)(b)=0, \text{ for all } b \in B\}\\ &=\{\sigma \in \text{Hom}_R(C,M) \mid \sigma(\beta(b))=0, \text{ for all } b \in B\} \\ &=\{\sigma \in \text{Hom}_R(C,M) \mid \sigma(c)=0 \text{ for all } c \in C\} \end{align} (because $$\beta$$ is surjective by supposition) $$=\{0\}.$$
Thus, $$\beta^{*}$$ is injective.

Now suppose that $$\sigma \in \text{Ker}(\alpha^{*})$$.
Then $$(\sigma \circ \alpha)(a)=0 \text{ for all } a \in A$$, implying that $$\text{Im}(\alpha) \subseteq \text{Ker}(\sigma)$$, which in turn implies that $$\text{Ker}(\beta) \subseteq \text{Ker}(\sigma)$$ (because the original sequence was exact).

Define a function $$\phi: C \to M$$ by the following:
For all $$c \in C$$, pick some $$b_c \in B$$ such that $$\beta(b_c)=c$$ (we know that $$\beta$$ is surjective from before).
Additionally, set $$\phi(c)=\sigma(b_c)$$.
Then because $$\sigma$$ is a homomorphism, then $$\phi$$ is also a homomorphism, meaning that $$\phi \in \text{Hom}_R(C,M)$$.
Let us consider the following
\begin{align} (\beta^{*}(\phi))(b_c)=(\phi \circ \beta)(b_c) &=\phi(\beta(b_c))\\ &=\phi(c)\\ &=\sigma(b_c). \end{align}
Therefore, $$\sigma \in \text{Im}(\beta^{*})$$, meaning that $$\text{Ker}(\alpha) \subseteq \text{Im}(\beta^{*})$$.

Now suppose that $$\sigma \in \text{Im}(\beta^{*})$$.
Then there must exist some $$\varphi \in \text{Hom}_R(C,M)$$ such that $$\beta^{*}(\varphi)=\varphi \circ \beta = \sigma$$.
We have the following:
$$\alpha^{*}(\sigma)=\sigma \circ \alpha = \varphi \circ \beta \circ \alpha = \varphi \circ 0 =0$$ (because $$\beta \circ \alpha =0$$).
Thus, $$\sigma \in \text{Ker}(\alpha^{*})$$, meaning that $$\text{Im}(\beta) \subseteq \text{Ker}(\alpha^{*})$$.
Thus, by double containment, we must have that $$\text{Ker}(\alpha^{*}) = \text{Im}(\beta^{*})$$, meaning that the sequence is exact, as was to be shown.

Any suggestions/feedback?

• Is $\phi$ really a homomorphism (for arbitrary choices of preimages)? Mar 27, 2020 at 0:52
• I believe it should be. Mar 27, 2020 at 0:59
• Yes, actually it is. The thing is that it is well defined (doesn't depend on the choices). Maybe you should prove that as well, and then your proof will be complete. Mar 27, 2020 at 1:10
• Thank you for your input! :) Mar 27, 2020 at 1:11

Generally looks good, though as Berci says in the comments you should check that $$\phi$$ doesn't depend on your choices.
However, I write an answer, because I would suggest an alternative method of proof for $$\newcommand\im{\operatorname{im}}\ker\alpha^*=\im\beta^*$$.
The key is to notice that $$C\cong B/\im\alpha$$, and $$\beta : B\to C$$ is the quotient map. Therefore you can use the universal property of the quotient, which is that maps $$\psi : B\to M$$ such that $$\psi(\im\alpha)=0$$ are in one-to-one correspondence with maps $$\tilde{\psi} : C\to M$$, and the correspondence is given by $$\psi = \tilde{\psi}\circ \beta$$.
Then $$\psi(\im\alpha)=0$$ if and only if $$\alpha^*\psi = \psi\circ \alpha =0$$ if and only if $$\psi\in \ker\alpha^*$$. Thus $$\psi\in\ker\alpha^*$$ if and only if $$\psi=\tilde{\psi}\circ \beta$$ for some $$\tilde{\psi}\in \operatorname{Hom}_R(C,M)$$, i.e., $$\psi \in \ker\alpha^*$$ if and only if $$\psi \in \im\beta^*$$, as desired.