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Let $\mathbf C$ be an abelian category containing arbitrary direct sums and let $\{X_i\}_{i\in I}$ be a collection of objects of $\mathbf C$.

Consider a subobject $Y\subseteq \bigoplus_{i\in I}X_i$ and put $Y_i:=p_i(Y)$ where $p_i:\bigoplus_{i\in I}X_i\longrightarrow X$ is the obvious projection.

Is $Y$ a subobject of $\bigoplus_{i\in I}Y_i$?

This seems so obvious, but I can't seem to be able to prove it.

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    $\begingroup$ Note that it might be tempting to prove a statement of this form using Freyd-Mitchell, but the embedding provided by Freyd-Mitchell is only guaranteed to be exact, and so is not guaranteed to commute with infinite coproducts. $\endgroup$ – Qiaochu Yuan Jan 20 '16 at 15:33
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Nope, this isn't true in general. For instance, let $\mathbf{C}=Ab^{op}$. Then you are asking the following question: given a surjective homomorphism $p:\prod X_i\to Y$ of abelian groups, let $K_i$ be the kernel of the restriction of $p$ to $X_i$. Then does $p$ factor through $\prod X_i/K_i$?

Of course, this is false, because (for instance) you could take $Y=\prod X_i/\bigoplus X_i$ and $p$ to be the quotient map, and then $K_i=X_i$ for all $i$, but $p$ does not factor through $\prod X_i/K_i=0$ as long as infinitely many of the $X_i$ are nontrivial. All you can say is that $p$ factors through $(\prod X_i)/(\bigoplus K_i)$, which is generally much larger than $\prod X_i/K_i=(\prod X_i)/(\prod K_i)$.

(In general, when you are having trouble proving some reasonable-seeming property of infinite direct sums or products, it is often a good idea to check whether it is true of categories like $Ab^{op}$, i.e. whether the dual statement is true in familiar abelian categories.)

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  • $\begingroup$ Okay, this may again be a stupid question, but how is it that the image corresponds to the kernel in the opposite category? Wouldn't it rather be the cokernel? $\endgroup$ – abeliancats Jan 20 '16 at 9:34
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    $\begingroup$ Image is actually self-dual (it is the kernel of the cokernel, which is the same as the cokernel of the kernel); the image is not $K_i$ but $X_i/K_i$. $\endgroup$ – Eric Wofsey Jan 20 '16 at 9:35
  • $\begingroup$ Wow...thanks! Do there exist conditions on the abelian category such as AB conditions that would allow something like this to hold? $\endgroup$ – abeliancats Jan 20 '16 at 10:35
  • $\begingroup$ Not that I know of off the top of my head, but it seems like it might follow from AB5. $\endgroup$ – Eric Wofsey Jan 20 '16 at 10:55
  • $\begingroup$ So we are good with Grothendieck categories? $\endgroup$ – abeliancats Jan 20 '16 at 11:07
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It's possible to be more concrete about Eric Wofsey's counterexample as follows. By Pontryagin duality, $\text{Ab}^{op}$ is equivalent to the category of compact abelian groups. What makes this category funny is that infinite direct sums are much bigger than you might naively think. The problem with naive infinite direct sums is, of course, that they are not compact. So you need to compactify them!

Explicitly, if $A_i$ is a collection of compact abelian groups, then their infinite direct sum $\bigoplus_i A_i$ is the Bohr compactification of the naive infinite direct sum. Said another way, it's the Pontryagin dual of $\prod_i A_i^{\vee}$, where $(-)^{\vee}$ denotes the Pontryagin dual. For example, if all of the $A_i$ are $S^1$, then $\bigoplus_i S^1$ is the Pontryagin dual of $\prod_i \mathbb{Z}$ (with the discrete topology). This is a huge group and it's much bigger than the naive infinite direct sum.

So intuitively, the reason $Y$ can fail to be a subobject of $\bigoplus_i Y_i$ is that Bohr compactification adds a lot of points, and Bohr compactifying this guy might not reach enough points to contain all of $Y$, which in general might intersect nontrivially with a whole bunch of extra points coming from the compactification.

Eric Wofsey's counterexample, translated into this language, says to take $Y$ to be the kernel of the map $\bigoplus_i S^1 \to \prod_i S^1$ (this map really does have a kernel! The direct sum is bigger than the direct product in this category!).

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