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

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.

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 $$\beta=f\circ\alpha=f_*(\alpha)$$.

• Ah yes I see, I knew that $x$ was unique but it was the observation that it's for all $y\in M$ that I wasn't thinking about. Thanks for the answer. Commented Jun 19, 2020 at 12:04