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I've just looked at the book "A Course in Homological Algebra" ( by Hilton and Stammbach) . They show the universal property of the direct sum (coproducts) using injections and the universal property of the direct product using projections. Then they prove that $\text{Hom}( \coprod_i A_i, B) \cong \prod_i \text{Hom} (A_i, B)$ and $\text{Hom}(A, \prod_i B_i) \cong \prod_i \text{Hom} (A, B_i)$. The question is why not use injections in the universal property of the coproduct of modules and why not $\text{Hom}( \prod_i A_i, B) \cong \prod_i \text{Hom} (A_i, B)$? I know it differs for general categories, because of the duality, but, for modules, isn't it possible to do this kind of generalisation?

Thanks in advance.

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It's a Goldilocks situation as the cnaonical injections and projections are just the right thing to prove the claims at hands. Moreover, direct sums and direct products also differ for modules.

Consider the case $B=A_i=\mathbb Z$ with index set $\mathbb N$. Then $\prod_{i} A_i$ is the set $\mathbb Z^{\mathbb N}$ of functions $\mathbb N\to \mathbb Z$.

What is $\prod_i \operatorname{Hom}(A_i,B)$? An element of this is determined by specifying for each $i\in \mathbb N$ an element of $B$ (the image of $1\in A_i$), hence "is" also $\mathbb Z^{\mathbb N}$ and countable.

What is $\operatorname{Hom}(\prod_i A_i,B)$? It is definitely bigger than $\mathbb Z^{\mathbb N}$ already because $\prod A_i$ is uncountable!

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If anyone reading this answer is surprised that the modules $\coprod_{i \in I}A_1$ and $\prod_{i \in I}A_i$ are not the same thing here is the likely source of your confusion: If the index set $I$ is finite then in fact they are the same! But when $I$ is an infinite set they are very different, as Hagen's example shows. –  Jim Feb 23 '13 at 16:10
    
$\mathbb{Z}^\mathbb{N}$ is not countable, but $Hom(\prod_\mathbb{N} \mathbb{Z},\mathbb{Z})$ is countable! –  Ralph Feb 23 '13 at 16:30
    
@Jim The source of my confusion wasn't in the fact that $\prod A_i$ is different from $\coprod A_i$. It was in the fact that $\prod$ cannot be pushed out from the $\text{Hom}$. –  user40276 Feb 23 '13 at 17:08
    
@Hagen von Eitzen So, that's not possible to define a universal property for product of modules using injections? Furthermore, if your index was not a limit ordinal, then the stament would hold, isn't? Thanks –  user40276 Feb 23 '13 at 17:23
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@user40276: A fact you might find interesting is the Yoneda lemma, which, neglecting some technicalities, says that if $\hom(A, M) \simeq \hom(B, M)$ holds for all $M$ then $A \simeq B$. So if $\hom(\prod A_i, M) = \prod\hom(A_i, M) = \hom(\coprod A_i, M)$ always held then we would have $\prod A_i \simeq \coprod A_i$. –  Jim Feb 23 '13 at 18:08
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Why not $\text{Hom}( \prod_i A_i, B) \cong \prod_i \text{Hom} (A_i, B)$ ?

$\text{Hom}( \prod_\mathbb{N} \mathbb{Z}, \mathbb{Z}) \cong \bigoplus_\mathbb{N}\mathbb{Z}$ is a free abelian group while $\prod_\mathbb{N} \text{Hom} (\mathbb{Z}, \mathbb{Z}) \cong \mathbb{Z}^\mathbb{N}$ isn't free. Hence they are not isomorphic.

As reference for the first isomorphism see http://mathoverflow.net/questions/105771/is-the-dual-of-the-product-of-infinite-cyclic-groups-a-free-abelian-group and http://www.math.uni-duesseldorf.de/~schroeer/publications_pdf/infinite_product-1.pdf for the latter.

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Thanks for the two links above. –  user40276 Feb 23 '13 at 17:31
    
I was wondering tho... if R is a field ... would it be true tho? –  CSA Feb 23 at 4:19
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