# What can $\operatorname{Hom}\left(\prod_i M_i, N\right)$ look like?

It's easy to see that $\operatorname{Hom}\left(\bigoplus_i M_i, N\right) = \prod_i \operatorname{Hom}(M_i, N)$. However, there are a couple of ways this can conceivably fail if we replace the coproduct on the left with a product: we could have a homomorphism $\bigoplus_i M_i \rightarrow N$ which didn't extend to $\prod_i M_i$, or we could have one that extended non-uniquely.

In the latter case, there would be a nonzero homomorphism $\prod_i M_i \rightarrow N$ which was identically zero on elements of "finite support." Can this happen, and if so is there a nice example? (Apologies if this is obvious -- I've been thinking about it for a while and can't come up with anything.)

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Well, by standard nonsense we have a homomorphism $\bigoplus_i \textrm{Hom}(M_i, N) \to \textrm{Hom}(\prod_i M_i, N)$. This corresponds to all the homomorphisms $\prod_i M_i \to N$ that factor through the product/sum of finitely many factors. To get a homomorphism that vanishes on the elements of "finite support", take $N = \operatorname{coker} (\bigoplus_i M_i \to \prod_i M_i)$. This is non-zero if and only if the product is strictly larger than the direct sum. – Zhen Lin Aug 8 '12 at 4:33

Here is a slightly simpler answer to your specific question. If your $M_i$ are modules over a ring, there is a canonical example of a homomorphism that kills the elements of finite support (which also works for many other algebraic structures): let $F \subseteq \prod_i M_i$ be the set of elements of finite support, then $F$ is a submodule; if you take $N$ to be the quotient module $(\prod_i M_i)/F$, then the projection $p : \prod_i M_i \rightarrow N$ is identically zero on $F$. Any homomorphism on $\prod_i M_i$ that maps $F$ to zero factors through $p$. It may help to think of $N$ as what you get from the product by identifying elements that are "almost identical", where $x$ and $y$ are "almost identical" if $x_i \not= y_i$ for only finitely many $i$.

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Wow, that was obvious. Some days I just miss things... – Daniel McLaury Aug 9 '12 at 4:51

By applying the hom functor to the exact sequence

$$0 \to \coprod_i M_i \to \prod_i M_i \to C \to 0$$

(where $C$ is defined as the cokernel), we get a long exact sequence

$$0 \to \mathop{\mathrm{Hom}}(C, N) \to \mathop{\mathrm{Hom}}\left(\prod_i M_i, N\right) \to \mathop{\mathrm{Hom}}\left(\coprod_i M_i, N\right) \\ \to \mathop{\mathrm{Ext^1}}(C, N) \to \mathop{\mathrm{Ext^1}}\left(\prod_i M_i, N\right) \to \mathop{\mathrm{Ext^1}}\left(\coprod_i M_i, N\right) \to \cdots$$

(the next term in the sequence is 0 if you're working with Abelian groups)

If you can understand the groups in this sequence, you can compute things. For example, understanding $C$ (and homomorphisms from $C$) tells you what homomorphisms $\prod_i M_i \to N$ vanish on elements of finite support. Or if $N$ is an injective module, then $\mathop{\mathrm{Ext^1}}(C, N) \cong 0$, and so every homomorphism $\coprod_i M_i \to N$ extends (probably not uniquely) to a homomorphism $\prod_i M_i \to N$. (this last statement, I believe, is not an 'if and only if')

(p.s. I really, really prefer limiting $\oplus$ to biproducts, and thus I don't like using it for infinite coproducts, which is why I use the notation $\coprod$ instead)

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I'm not sure whether nonzero homomorphisms $\prod_i M_i\to N$ can exist which are trivial on elements of finite support, but there are certainly homomorphisms $\bigoplus_i M_i\to N$ which do not extend to $\prod_i M_i\to N$. Namely, the homomorphism $\bigoplus_{i=1}^\infty M\to M$ given by $(m_1,m_2,\ldots)\mapsto \sum\limits_{i=1}^\infty m_i$ does not extend to $\prod_{i=1}^\infty M$.

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If $M$ is injective it does, because of the exact sequence $$0 \to hom( \prod_i M / \coprod_i M, M) \to hom(\prod_i M, M) \to hom(\coprod_i M, M) \to ext(\prod_i M / \coprod_i M, M)$$ The extensions simply aren't given by that formula. – Hurkyl Aug 8 '12 at 4:38
@Hurkyl Hmm, hadn't considered that. But certainly if we take something like $M=\mathbb Z_2$ then it won't extend. – Alex Becker Aug 8 '12 at 4:53
$\mathbb{Z}_2$ is the 2-adics? Can you describe the argument that it doesn't extend? Alas I've always had trouble working with the product modulo the coproduct so it's not clear to me. (If you meant the group with 2 elements, my same argument applies, just done in the category of $\mathbb{F}_2$-vector spaces) – Hurkyl Aug 8 '12 at 5:03
@Hurkyl I mean the $\mathbb Z$-module $\mathbb Z_2$. Working on the formal argument, give me a minute. – Alex Becker Aug 8 '12 at 5:05