$\hom_A(-,N)$Functor Takes Coproducts to Products

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1. Jul 19, 2015

NihilTico

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 $\bf{Ab}$.

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 $I$ 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:

$\begin{array}{cccc} & \jmath_{i} & \bigoplus M_{i}\\ M_{i} & \nearrow & \downarrow\\ & \searrow & \ \downarrow h\\ & f_{i} & C \end{array}$

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 $q\colon\hom_{A}\left(\bigoplus M_{i},N\right)\to\prod\hom_{A}\left(M_i,N\right)$.

2. Relevant equations

Not applicable.

3. The attempt at a solution

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

2. Jul 20, 2015

micromass

All of what you wrote is correct.

3. Jul 20, 2015

NihilTico

Thanks for the verification. What does this say about hom-functors? Is there a word for this property?

4. Jul 20, 2015

micromass

I would say that $\text{Hom}(-,N): \textbf{Mod}^\text{op}\rightarrow \textbf{Ab}$.preserves products.

5. Jul 20, 2015