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

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SUMMARY

The discussion centers on the left-exact functor \hom_{A}(-,N) as described in Lang's Algebra (3rd edition). It establishes that \hom_{A}\left(\bigoplus M_{i},N\right) is isomorphic to \prod\hom_{A}\left(M_i,N\right), where M_i and N are A-modules. The conversation emphasizes the unique determination of morphisms in this context, highlighting the relationship between coproducts and products in the categories \text{Mod}(A) and \bf{Ab}. The participants confirm the correctness of these observations and inquire about deeper implications regarding hom-functors.

PREREQUISITES
  • Understanding of left-exact functors in category theory
  • Familiarity with coproducts and products in the categories \text{Mod}(A) and \bf{Ab}
  • Knowledge of A-modules and A-homomorphisms
  • Basic concepts from Lang's Algebra, specifically the third edition
NEXT STEPS
  • Study the properties of left-exact functors in category theory
  • Explore the relationship between coproducts and products in \text{Mod}(A)
  • Investigate the implications of hom-functors in Mac Lane's "Categories for the Working Mathematician"
  • Examine the concept of isomorphisms in the category \bf{Ab}
USEFUL FOR

Mathematicians, particularly those specializing in category theory, algebraists working with A-modules, and students seeking to deepen their understanding of hom-functors and their properties.

NihilTico
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A couple of notes first:
1.
\hom_{A}(-,N) 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 \bigoplus M_i lies in the category \text{Mod}(A) while the product \prod\hom_{A}\left(M_i,N\right) lies in \bf{Ab}.

Homework Statement



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 I unless I feel it is necessary for clarity) what generalizes naturally to \hom_{A}\left(\bigoplus M_{i},N\right)\approx\prod\hom_{A}\left(M_i,N\right). Where the M_i and N are $A$-modules. But he goes about demonstrating it for the case of two A-modules, curiously.

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

<br /> \begin{array}{cccc} &amp; \jmath_{i} &amp; \bigoplus M_{i}\\ M_{i} &amp; \nearrow &amp; \downarrow\\ &amp; \searrow &amp; \ \downarrow h\\ &amp; f_{i} &amp; C \end{array}

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

Homework Equations


[/B]
Not applicable.

The Attempt at a Solution


[/B]
Not so much as an attempt, as an observation.

Clearly, any h\colon\bigoplus M_i\to{N} induces a unique family of A-homomorphisms given by \left\{h_i\colon M_i\to{N}\right\}_{i\in{I}} ; after all, if another family had this h 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 h from the coproduct to N. Isomorphisms in \bf{Ab} are bijective so this is enough by defining q(h)=\left(h_i\right)_{i\in{I}}.

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
 
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All of what you wrote is correct.
 
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micromass said:
All of what you wrote is correct.
Thanks for the verification. What does this say about hom-functors? Is there a word for this property?
 
I would say that ##\text{Hom}(-,N): \textbf{Mod}^\text{op}\rightarrow \textbf{Ab}##.preserves products.
 
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Thanks for your help!
 

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