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Demonstrating that a mapping is injective

  1. May 13, 2014 #1
    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, [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]


    [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.

  2. jcsd
  3. May 15, 2014 #2


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    Homework Helper

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

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