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I have a functor $F\colon \mathbf{Rng}\to\mathbf{Grp}$, and a correspondence on objects which assigns to every group $F(R)$ a suitable subgroup $G_o(R)\subseteq F(R)$. Is there a way to turn $G$ into a functor, defining $G_o(R)\to G_o(S)$ via the maps I have between $F(R)$ and $F(S)$? $$ \begin{array}{ccc} F(R) &\to^{F(f)}& F(S) \\\ \uparrow_{\iota_R} && \uparrow_{\iota_S}\\\ G_o(R) & &G_o(S) \end{array} $$ In this diagram vertical arrows are simply the existing injections. I thought to define $G_o(R)\to G_o(S)$ taking the obvious left inverse going from the copy of $G_o(-)$ into $F(-)$ to $G_o(-)$, (call $\pi_S$ this map, thn $G(R)\to G(S)$ is defined by $\pi_s\circ F(f)\circ \iota_R$) but I'm not sure this is going to work...

If you think you'll find it useful, $F$ is the functor which assigns to every ring its group of unities.

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Those injections are not morphisms in the category of groups, or rings, just sets. I think that is where you are going to bump into a problem. More importantly, what do you want out of this functor? Do you want this process to be functorial for a reason? – BBischof Dec 25 '10 at 16:13
Can we ask what exactly $G_o$ is? – Sean Tilson Dec 25 '10 at 16:17
I think usually a subfunctor is defined as already being a functor (so the maps you want to be there are part of the definition), such that its values are subobjects of the values of the big functor. – Dylan Wilson Dec 25 '10 at 18:35
up vote 1 down vote accepted

Using only the things you have written, your question does not have an answer. In general there may not be any way to complete the commutative diagram you drew into a square (and in general there may not be any projection $\pi_S$, but this is a smaller problem). The basic problem is that you have only specified what $G_0$ does on objects, so as it stands $G_0$ is not even a functor (and in particular not a subfunctor).

Consider the example where $F$ is the group of units functor, as above. Let $G_0(R)$ be the whole group of units whenever $R$ is infinite, and the trivial group when $R$ is finite. Then you will see that there is in general no way to fill in the square above for e.g. the map $\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$.

Fortunately the condition that $G_0$ should be a subfunctor of $F$ means that you have very little choice in how you define $G_0(f)$, for a morphism $f$. The fact that the diagram above should be commutative forces you to choose $G_0(f)$ to be equal to the restriction of $F(f)$ to $G_0(R)$. What you then need to check, in order for this to be well defined, is that this restriction always lands inside $G_0(S)$. (This fails in the example of the previous paragraph.) Whenever this condition holds, $G_0$ automatically becomes a functor using only that $F$ is a functor (check this).

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This was my original idea: I want to fix an ideal $J$ in a ring $R$, and then define the set of all subgroups $G\le R^\times$ such that $J+G=R^\times$. Call $S_J(R)$ the intersection of all such groups. $G_o$ is the correspondence associating $R$ to $S_J(R)$. I'm wondering if, given a ring morphism $f:R\to S$, i can obtain a morphism between $S_J(R)$ and $S_{f(J)}(S)$, where $f(J)$ is the ideal generated by $f(J)$. – Fosco Loregian Dec 26 '10 at 16:33

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