Demonstrating that a mapping is injective

[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

 

[pasmith]

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?

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.