# $\hom_A(-,N)$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

Staff Emeritus
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

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

5. Jul 20, 2015