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Background / Motivation:

Consider the functor $S \colon \mathrm{Mod} \to \mathrm{Comm.Alg}$ which sends a module to the symmetric algebra over that module. Let $M$ be a $k$-module and let $R \subseteq M$ be a set of relations in $M$. Then we have

$S(M)/(R) = S(M/<R>)$

via the isomorphism defined on generators by: $m + (R) \mapsto m + \langle R\rangle$

Hence we get the same result whether we add in relations at the module level, or at the algebra level.

We also have the result in the reverse direction taking the right adjoint to $S$ which is the functor which sends $A$ to it's irreducible elements.

The Problem

I'm hoping to generalize this result to different categories (in particular a specific set which have some nice properties).

I have a 'proof' of such a result, although I fall down when trying to phrase it in a category theoretical context. There are two particular issues I am having, which I believe are related.

The context is given sufficiently 'nice' (concrete, isomorphism theorems etc.) categories $C, D$, and a pair of adjoint functors $F \colon C \to D $, $G \colon D \to C$, take an object $A \in C$ and a set of relations $R \subseteq A$. N.B. $G$ is not necessarily forgetful.

Problem 1: I would like there to be a way of viewing $A$ (and hence the relations) as a subset of $FA$. The unit of our adjunction gives a map $A \to GFA$. However $GX$ is not necessarily a subset of $X$ (although I can't think of any counter examples to this?)

Problem 2: In the course of the proof, I would like that given a map in $C$, $f \colon A \to B$, then $Ff \colon FA \to FB$ should restrict to $f$ on the subset $A$ inside $FA$. I'm unclear what this means categorically...

Any help or ideas on how to proceed would be greatly appreciated, been stuck on this for a week now! I think once I get the correct categorical setting I should be home and dry, but have been unable to do this.

Thanks

Will

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  • $\begingroup$ You might like to start by considering the consequences of having the right adjoint be faithful. $\endgroup$ – Zhen Lin Nov 4 '14 at 20:36
  • $\begingroup$ Thanks for the suggestion, I don't quite see how that would help in this context however? Did you have a specific idea in mind as to where it can be used? $\endgroup$ – WMycroft Nov 4 '14 at 21:06
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    $\begingroup$ Solution of Problem 1: Don't forget forgetful functors! The map $A \to GFA$ is the "inclusion map". Don't confuse $GFA$ with $FA$. Solution of Problem 2 follows. $\endgroup$ – Martin Brandenburg Nov 5 '14 at 8:03
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The general statement is that left adjoints are cocontinuous: that is, they preserve colimits. Taking a quotient of an object by relations is a special case of a colimit: it's a coequalizer.

Let me start by describing the construction for $\text{Set}$. If $X$ is a set and $\sim$ is an equivalence relation on $X$, let

$$R = \{ (x, y) \in X \times X : x \sim y \}$$

together with the two projection maps $r_1, r_2 : R \to X$. Then my claim is that the quotient of $X$ by the equivalence relation $\sim$ is precisely the coequalizer of $r_1$ and $r_2$.

This works in great generality. In your case, if $M$ is a module and $N$ a submodule you would like to quotient out by, let

$$R = \{ (m, m') \in M \times M : m \equiv m' \bmod N \}.$$

This is a submodule of $M \times M$, and in particular a module, equipped with two projection maps $r_1, r_2 : R \to M$. Now my claim is that the quotient of $M$ by $N$ is precisely the coequalizer of $r_1$ and $r_2$, and the corresponding statement for the quotient of $\text{Sym}(M)$ by the relations imposed by $N$ also holds.

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  • $\begingroup$ Hi Qaiochu, thanks for this. I think you are right in that coequalizers are the best way to approach this problem. However, I don't think this is the whole story. The remaining question is what we mean by the ideal/submodule generated by R (recall R is a subset of A) in a general category. Obviously we won't always be able to define such an object, but I was wondering under which conditions is it possible. $\endgroup$ – WMycroft Nov 5 '14 at 17:35
  • $\begingroup$ @WMycroft: this approach avoids having to answer that question. Our functor from modules to algebras just converts a parallel pair of arrows $r_1, r_2 : R \to M$ to another parallel pair of arrows $S(r_1), S(r_2) : S(R) \to S(M)$. (In category theory it's generally unnatural to impose extra conditions like that various morphisms are monomorphisms, e.g. because those conditions 1) often turn out to be unnecessary and 2) often aren't preserved under the functors we care about.) $\endgroup$ – Qiaochu Yuan Nov 5 '14 at 20:12
  • $\begingroup$ I agree in general this would be unnatural and probably impossible. However I'm working with some categories with very nice properties, they are all various types on algebraic objects in the category of modules, so I can afford to lose a lot of generality in the hope of achieving my desired statement. $\endgroup$ – WMycroft Nov 5 '14 at 21:53
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Let $U : \mathsf{CAlg}(k) \to \mathsf{Mod}(k)$ be the forgetful functor, so that the symmetric algebra construction $S$ is left adjoint to $U$. If $A$ is a commutative $k$-algebra, then there are natural bijections $$\hom(S(M)/ \langle R \rangle ,A) \cong \{f \in \hom(S(M),A) : R \subseteq \ker(f)\}$$ $$ \cong \{g \in \hom(M,U(A)) : R \subseteq \ker(g)\} \cong \hom(M/\langle R \rangle,A)$$ and hence $S(M) / \langle R \rangle \cong M/\langle R \rangle$.

Now let us generalize the setting:

Let $C,D$ be arbitrary categories. Let $U : D \to C$ be a functor which has a left adjoint $S : C \to D$. Let $f,g : N \rightrightarrows M$ be two morphisms in $C$ and let $M \to Q$ be their coequalizer (assume it exists), i.e. the universal quotient of $M$ which imposes the relation $f=g$ on $Q$. Since $S$ is left adjoint and hence cocontinuous, $S(M) \to S(Q)$ is the coequalizer of $S(f),S(g) : S(N) \rightrightarrows S(M)$. That is, if $A \in D$, then morphisms $S(M) \to A$ correspond to morphisms $S(M) \to A$ which coequalize $S(f),S(g) : S(N) \rightrightarrows S(M)$. But since $S$ is left adjoint to $N$, this precisely means that $M \to U(S(M)) \to U(A)$ coequalizes $f,g : N \rightrightarrows M$. This may be imagined as the statement that $S(Q)$ is universal quotient of $S(M)$ which just imposes the relation $f=g$ on the "underlying" objects in $C$ (with respect to $U$).

For example, let $U : \mathsf{CAlg}(k) \to \mathsf{Mod}(k)$ be the forgetful functor, and let $N$ be a submodule of $M$. Let $f$ be the inclusion and let $g$ be the zero map, so that $Q$ is the usual quotient module $M/N$. The argument above shows that $S(M/N)$ is the universal quotient of $S(M)$ which makes the elements of $N$ or rather the map $N \to U(S(M)) \to U(S(M/N))$ to zero. Equivalently, if $\langle N \rangle$ denotes the ideal generated by (the underlying set of) $N$ in $S(M)$, then $S(M/N)=S(M)/\langle N \rangle$.

I think that you would like to generalize the last step, namely the construction of $\langle N \rangle$, for arbitrary categories. I don't think that this works in general. The main problem is that ideals are no subalgebras, so that we leave our category. But in the setting of non-unital algebras, this problem disappears. Notice the category of non-unital algebras has a zero object. We can give now the following generalization:

Let $U : D \to C$ be as above, but assume that $C,D$ have zero objects which are preserved by $S,U$. Let $N \to M$ be just one morphism in $C$. Let $M/N$ be the cokernel in $C$ (i.e. the coequalizer of $N \to N$ and the zero morphism). Let $S(N) \to \langle N \rangle \to S(M)$ be an epi-mono factorization (assume it exists; it happens in most cases). Then $S(M/N)=S(M)/\langle N \rangle$.

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  • $\begingroup$ Hi Martin, thanks for your response, I think this is exactly what I need. There's one bit in particular I'm struggling to follow, could you explain this step: "But since S is left adjoint to U, this precisely means that M→U(S(M))→U(A) coequalizes f,g:N⇉M" I see where the map arises from, the adjunction unit followed by the underlying functor applied to our morphism from S to A. However I'm unclear as to why this coequalizes f and g. $\endgroup$ – WMycroft Nov 5 '14 at 18:13
  • $\begingroup$ Have you tried using $\hom(N,U(A)) \cong \hom(S(N),A)$? $\endgroup$ – Martin Brandenburg Nov 5 '14 at 20:27
  • $\begingroup$ I see why that might be useful, we can apply it to our two maps (one using f, the other g) N→M→U(S(M))→U(A), to yield corresponding maps S(N)→A. But why are these corresponding maps equal? $\endgroup$ – WMycroft Nov 5 '14 at 21:46

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