I am reading Dummit and Foote, Section 10.5 : Exact Sequences - Projective, Injective and Flat Modules.(adsbygoogle = window.adsbygoogle || []).push({});

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, [itex] Hom_R (R, X) \cong X [/itex], the isomorphism being given by mapping a homomorphism to its value on the element [itex]1 \in R [/itex]"

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

To demonstrate the isomorphism I proceeded as follows:

Let [itex]f, g \in Hom_R (R,X) [/itex] so [itex] f,g : \ R \to X [/itex]

Consider [itex] \theta \ : \ Hom_R (R,X) \to X [/itex]

where [itex] \theta (f) = f(1_R) [/itex]

To show [itex] \theta [/itex] is a homomorphism we proceed as follows:

[itex] \theta (f + g) = (f + g)(1_R) = f(1_R) + g(1_R) [/itex]

[itex] = \theta (f) + \theta (g) [/itex]

and

[itex] \theta (rf) = (rf) = rf (1_R) = r \theta (f)[/itex] where [itex] r \in R [/itex]

Then I need to show [itex] \theta [/itex] is injective and surjective.

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

Note that I suspect one would proceed as follows:

Suppose we have [itex] f, g \in Hom_R (R,X) [/itex] such that:

[itex] \theta (f) = f(1_R) [/itex] and [itex] \theta (g) = f(1_R) [/itex]

Now we have, of course, by definition of g, that [itex] \theta (g) = g(1_R) [/itex]

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

Hope someone can help.

Peter

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# Demonstrating that a mapping is injective

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