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[itex]\hom_A(-,N)[/itex]Functor Takes Coproducts to Products

  1. Jul 19, 2015 #1
    A couple of notes first:
    1.
    [itex]\hom_{A}(-,N)[/itex] is the left-exact functor I'm referring to; Lang gives an exercise in the section preceeding to show this.
    2.
    This might be my own idiosyncrasy but I write TFDC to mean 'The following diagram commutes'
    3.
    Titles are short, so I know that the hom-functor here isn't actually taking a coproduct to a product in the same category. The coproduct [itex]\bigoplus M_i[/itex] lies in the category [itex]\text{Mod}(A)[/itex] while the product [itex]\prod\hom_{A}\left(M_i,N\right)[/itex] lies in [itex]\bf{Ab}[/itex].

    1. The problem statement, all variables and given/known data

    I don't think I saw this before, but on page 131 of Lang's Algebra (3rd edition) he writes (NOTE: I'm dropping the indexing set [itex]I[/itex] unless I feel it is necessary for clarity) what generalizes naturally to [itex]\hom_{A}\left(\bigoplus M_{i},N\right)\approx\prod\hom_{A}\left(M_i,N\right)[/itex]. Where the [itex]M_i[/itex] and [itex]N[/itex] are $A$-modules. But he goes about demonstrating it for the case of two [itex]A[/itex]-modules, curiously.

    To be a thorough as possible, is it the case, since the coproduct is the initial object in the category of tuples [itex]\left(C,\left\{f_{i}\colon{M_i}\to{C}\right\}_{i\in{I}}\right)[/itex] (where the [itex]f_i[/itex] are [itex]A[/itex]-homomorphisms) meaning that a morphism [itex]f\colon\left(\bigoplus M_i,\left\{\jmath_{i}\colon{M_i}\to{\bigoplus {M_i}}\right\}_{i\in{I}}\right)\to\left(C,\left\{f_{i}\colon{M_i}\to{C}\right\}_{i\in{I}}\right)[/itex] is a unique [itex]A[/itex] homomorphism [itex]h\colon \bigoplus M_i\to{C}[/itex] such that [itex]\forall i\in{I}[/itex] TFDC:

    [itex]
    \begin{array}{cccc} & \jmath_{i} & \bigoplus M_{i}\\ M_{i} & \nearrow & \downarrow\\ & \searrow & \ \downarrow h\\ & f_{i} & C \end{array}[/itex]

    that this is totally obvious, since [itex]h[/itex] is uniquely determined by the family [itex]\left\{f_i\right\}_{i\in{I}}[/itex]? In particular, the problem becomes—more or less—one of the existence of a group homomorphsim [itex]q\colon\hom_{A}\left(\bigoplus M_{i},N\right)\to\prod\hom_{A}\left(M_i,N\right)[/itex].

    2. Relevant equations

    Not applicable.

    3. The attempt at a solution

    Not so much as an attempt, as an observation.

    Clearly, any [itex]h\colon\bigoplus M_i\to{N}[/itex] induces a unique family of [itex]A[/itex]-homomorphisms given by [itex]\left\{h_i\colon M_i\to{N}\right\}_{i\in{I}}[/itex] ; after all, if another family had this [itex]h[/itex] as well, then the families would be equal as well by definition of the coprouct. Similarly, for any family in the product on the right, there is a unique [itex]h[/itex] from the coproduct to [itex]N[/itex]. Isomorphisms in [itex]\bf{Ab}[/itex] are bijective so this is enough by defining [itex]q(h)=\left(h_i\right)_{i\in{I}}[/itex].

    Right?

    If the above is correct, does it speak to something deeper about hom-functors that I'd be better served finding in Mac Lane's CWM?


    Thanks
     
  2. jcsd
  3. Jul 20, 2015 #2

    micromass

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    All of what you wrote is correct.
     
  4. Jul 20, 2015 #3
    Thanks for the verification. What does this say about hom-functors? Is there a word for this property?
     
  5. Jul 20, 2015 #4

    micromass

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    I would say that ##\text{Hom}(-,N): \textbf{Mod}^\text{op}\rightarrow \textbf{Ab}##.preserves products.
     
  6. Jul 20, 2015 #5
    Thanks for your help!
     
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