Proof that $\text{Hom}_R(M, -)$ is left exact in the category of $R$-modules

Question

I'm looking at a proof that $\text{Hom}_R(M, -)$ is left exact for $R$-modules. Specifically at the one that appears in Robert Ash's Abstract Algebra, which you can find here on page 13.

Let $A, B, C$ be $R$-modules for commutative ring $R$ and suppose

$$0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0 $$

is a short exact sequence. And consider

$$ 0\to \text{Hom}_R(M, A) \xrightarrow{f_\ast} \text{Hom}_R(M, B) \xrightarrow{g_*} \text{Hom}_R(M, C)$$

I understand everything in Ash's proof except for the very last step in proving that $\ker{g_*}\subseteq \text{im}f_*$.

Suppose $\beta\in\ker{g_*}$, then $g\circ\beta = 0$, and therefore for some $y\in M$ we have $g(\beta(y)) = 0$. So $\beta(y)\in\ker{g}=\text{im}f$. Therefore there is some $x\in A$ such that $\beta(y) = f(x)$. Here is where I have issues. Ash states that $x = \alpha(y)$ for $\alpha\in\text{Hom}_R(M, A)$. But how can one be certain that such a homomorphism exists?

The answer that appears here suffers from a similar issue. Here a function $l:M\to A$ is defined such that $l(y) = x$, but it isn't shown to be a homomorphism and I'm not certain how you would show that, if it's even possible from such a definition.

Answer 1

First of all, $g\circ\beta=0$ means for all $y\in M$, $g(\beta(y))=0$ (not only for some).

So let $y\in M$. Since $\beta(y)\in\ker(g)=im(f)$, there exists $x\in A$ such that $\beta(y)=f(x)$. The crucial point that you're missing is this one: since $f$ is injective by assumption, this $x$ is unique!! (if $\beta(y)=f(x_1)=f(x_2)$, then $x_1=x_2$...)

Hence, we may denote by $\alpha(y)$ this $x$, and we get a map $\alpha:M\to A$ such that $\beta(y)=f(\alpha(y))$ for all $y\in M$. Thus, $\beta=f\circ \alpha$.

Now it remains to prove that $\alpha$ is $R$-linear. By definition, for $y\in M$, $\alpha(y)$ is the unique element of $M$ such that $\beta(y)=f(\alpha(y))$.

But for all $y_1,y_2\in M$ and all $r\in R$, we have $\beta(y_1+ry_2)=\beta(y_1)+r\beta(y_2)=f(\alpha(y_1))+r f(\alpha(y_2))=f(\alpha(y_1)+r\alpha(y_2))$. But uniqueness above, $\alpha(y_1+ry_2)=\alpha(y_1)+r\alpha(y_2)$, and we are done.

Since $\alpha$ is $R$-linear we have $\beta=f\circ\alpha=f_*(\alpha)$.