# Demonstrating that a mapping is injective

1. May 13, 2014

### Math Amateur

I am reading Dummit and Foote, Section 10.5 : Exact Sequences - Projective, Injective and Flat Modules.

I am studying Proposition 28 (D&F pages 387 - 388)

In the latter stages of the proof of Proposition 28 we find the following statement (top of page 388):

"In general, $Hom_R (R, X) \cong X$, the isomorphism being given by mapping a homomorphism to its value on the element $1 \in R$"

I am having some trouble in demonstrating the isomorphism involved in the relationship $Hom_R (R, X) \cong X$.

To demonstrate the isomorphism I proceeded as follows:

Let $f, g \in Hom_R (R,X)$ so $f,g : \ R \to X$

Consider $\theta \ : \ Hom_R (R,X) \to X$

where $\theta (f) = f(1_R)$

To show $\theta$ is a homomorphism we proceed as follows:

$\theta (f + g) = (f + g)(1_R) = f(1_R) + g(1_R)$

$= \theta (f) + \theta (g)$

and

$\theta (rf) = (rf) = rf (1_R) = r \theta (f)$ where $r \in R$

Then I need to show $\theta$ is injective and surjective.

BUT ... I am having problems in demonstrating that $\theta$ is injective ... can someone help me with this task?

Note that I suspect one would proceed as follows:

Suppose we have $f, g \in Hom_R (R,X)$ such that:

$\theta (f) = f(1_R)$ and $\theta (g) = f(1_R)$

Now we have, of course, by definition of g, that $\theta (g) = g(1_R)$

So $f(1_R) = g(1_R)$ ... but how do we proceed from here to show that f = g?

Hope someone can help.

Peter

2. May 15, 2014

### pasmith

You haven't yet used the fact that $f$ and $g$ are not arbitrary functions from $R$ to $X$, but morphisms in the category of R-modules.

If $f \in \mathrm{Hom}_R(X_1, X_2)$ then by definition $f$ is an additive-group homomorphism which also satisfies $$f(rx) = rf(x)$$ for all $r \in R$ and all $x \in X_1$.