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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
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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 $f(J)$. –  tetrapharmakon Dec 26 '10 at 16:33
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