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

In summary, the author demonstrates that if the coproduct is the initial object in a category, then there is a unique homomorphism from the coproduct to the category.
  • #1
NihilTico
32
2
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].

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 [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].

Homework Equations


[/B]
Not applicable.

The Attempt at a Solution


[/B]
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
 
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  • #2
All of what you wrote is correct.
 
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  • #3
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?
 
  • #4
I would say that ##\text{Hom}(-,N): \textbf{Mod}^\text{op}\rightarrow \textbf{Ab}##.preserves products.
 
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  • #5
Thanks for your help!
 

What is the [itex]\hom_A(-,N)[/itex] functor?

The [itex]\hom_A(-,N)[/itex] functor is a mathematical tool used in category theory to map objects in a category to sets, and morphisms to functions between these sets. It takes an object N in a category A and maps it to the set of all morphisms from any object in A to N.

What does it mean for a functor to take coproducts to products?

A functor that takes coproducts to products is one that preserves the structure of coproducts in a category. This means that if an object A is the coproduct of two objects B and C, the functor will map this coproduct to the product of the functor applied to B and the functor applied to C.

What is the significance of a functor taking coproducts to products?

A functor that takes coproducts to products is important because it helps to show the relationship between different objects in a category. It allows us to study the properties of a category by looking at the structure of its coproducts and products, and how they are preserved by the functor.

What are some examples of functors that take coproducts to products?

One example of a functor that takes coproducts to products is the dual functor, which maps an object A to its dual space A*, and maps a morphism f to its dual f* defined as the transpose of f. Another example is the tensor product functor, which takes two vector spaces and maps them to their tensor product, preserving the structure of the coproducts.

What is the relationship between coproducts and products in category theory?

In category theory, coproducts and products are dual concepts, meaning that they are related in a symmetric way. Coproducts are defined as the categorical sum of two objects, while products are defined as the categorical product of two objects. The fact that the [itex]\hom_A(-,N)[/itex] functor takes coproducts to products is an example of this duality, showing the relationship between these two concepts.

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