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?

  • 1
    $\begingroup$ Is $\phi$ really a homomorphism (for arbitrary choices of preimages)? $\endgroup$
    – Berci
    Mar 27, 2020 at 0:52
  • $\begingroup$ I believe it should be. $\endgroup$
    Mar 27, 2020 at 0:59
  • 2
    $\begingroup$ 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. $\endgroup$
    – Berci
    Mar 27, 2020 at 1:10
  • $\begingroup$ Thank you for your input! :) $\endgroup$
    Mar 27, 2020 at 1:11

1 Answer 1


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.


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