# The Unapologetic Mathematician

## The First Isomorphism Theorem (for Abelian Categories)

We had versions of the first isomorphism theorem for groups, rings, and modules. Now we’ll do it in a more general setting. We’re going to use it in our study of abelian categories, but it works in all three of the above cases.

As we said last time we talked about abelian categories, we really just use that our categories all have a zero object, kernels, and cokernels. We’ll work with that for now, and note that these properties hold in particular for abelian categories.

Now, if $\mathcal{C}$ is a category with a zero object, kernels, and cokernels, then any arrow $f$ of $\mathcal{C}$ has a factorization $f=m\circ q$, where $m=\mathrm{Ker}(\mathrm{Cok}(f))$. This holds because $\mathrm{Cok}(f)\circ f=0$, and by the universal property of kernels there is a unique $q$ with $f=m\circ q$.

On the other hand, if we have another factorization $f=m'\circ q'$ with $m'$ also a kernel, then we have a unique $t$ so that the following diagram commutes.

That is, the canonical factorization described above is in some sense the universal such factorization. Furthermore, if $\mathcal{C}$ has equalizers and every monic in $\mathcal{C}$ is a kernel, then $q$ is epic. In particular, these hypotheses are all satisfied for abelian categories. This is the isomorphism theorem — that every arrow factorizes essentially uniquely as the composition of an epic $q$ and a monic $m$.

We prove this by considering the following diagram:

We draw $p'=\mathrm{Cok}(m')$. Since $m'$ is itself a kernel, we see that $m'=\mathrm{Ker}(p')$. We also draw $p=\mathrm{Cok}(m)=\mathrm{Cok}(f)$. Now $p'\circ m'=0$, so $p'\circ f=p'\circ m'\circ q'=0$. By the universal property of $p$ , there is a unique $w$ so that $p'=w\circ p$. And then $p'\circ m=w\circ p\circ m=0$, so $m$ factors uniquely through $m'$ — there is a unique $t$ with $m=m'\circ t$. Furthermore, $m'\circ q'=m\circ q=m'\circ t\circ q$, and so since $m'$ is monic we have $q'=t\circ q$. This proves that the first diagram given above commutes.

Now we have to show that $q$ is epic under the additional hypotheses. Let’s say that we have a parallel pair of arrows $r$ and $s$ with $r\circ q=s\circ q$. then $q$ factors uniquely through the equalizer $\mathrm{Equ}(r,s)$$q=e\circ q'$ for some unique $q'$. Then $f=m\circ q=m\circ e\circ q'$. Now $m'=me$ is monic, so (by the new hypotheses) it’s a kernel. By the first part of the theorem we have a unique $t$ with $m=m'\circ t=m\circ e\circ t$, and thus $1=e\circ t$. Since the monic $e$ has a right inverse, it’s an isomorphism. Since $e$ was picked as the equalizer of $r$ and $s$, $r=s$. And so $q$ is epic.

September 25, 2007 Posted by | Category theory | 8 Comments

## Problems with Ordered Linear Spaces — Solved!

Okay, there’s some sort of problem with these things. I defined the category $\mathcal{O}rd\mathcal{L}in$, showed some properties, and tried to prove a theorem. Here I want to collect what I’m sure about and what I’m not sure about.

The biproduct, exponential, and monoidal product structures are all correct. We have something natural defined by $\hat{V}=\hom_{\mathcal{O}rd\mathcal{L}in}(V,\mathbb{F})$, which seems sort of like a categorical dual, but maybe not.

What I tried to prove is that in general $\hat{V}\otimes W$ is not isomorphic to $W^V$. But as Todd pointed out, if $\hat{V}$ is a dual then we have an adjunction
$V\otimes\underline{\hphantom{X}}\dashv\hat{V}\otimes\underline{\hphantom{X}}$
but the definition of an exponential is an adjunction
$V\otimes\underline{\hphantom{X}}\dashv(\underline{\hphantom{X}})^V$
and functors with the same left adjoint are naturally isomorphic.

So, if $\hat{V}$ is a categorical dual, then $W^V$ is naturally isomorphic to $\hat{V}\otimes W$, and there’s something wrong with the proof I offered that it isn’t.

On the other hand, if $\hat{V}\otimes W$ is not naturally isomorphic to $W^V$, then $\hat{V}$ is not a categorical dual. I haven’t found an example of a cone that has no positive basis with a positive dual basis, and I’m losing faith in the example I thought I’d sketched that even had one positive basis whose dual basis was not positive.

So one of these “proofs” is wrong — which is it?

[UPDATE]: Well, I figured it out. There’s a problem with the duals.

Start with any cone $V$, and pick a positive basis in the cone. Now we can simplify our lives by moving to a cone with a particularly nice positive basis. We know that there’s a linear isomorphism $g$ taking the vector space $V$ to the free vector space $\mathbb{F}^n$, and taking our chosen basis of $V$ to the canonical basis of $\mathbb{F}^n$. We induce a cone on $\mathbb{F}^n$ by saying $(x_1,...,x_n)\geq0$ if and only if $g^{-1}(x_1,...,x_n)\geq_V0$. So without loss of generality our positive basis consists of the “coordinate vectors” $(0,...,0,1,0,...,0)$, and the dual basis consists of the coordinate projections.

Now these vectors cut out a classical subcone of $V$, and the positive vectors are exactly those in the positive orthant. If $V$ is not exactly this classical cone, then there exists some positive vector $v$ in $V$ outside the positive orthant. That is, some component of $v$ will be negative. But then that coordinate projector will send $v$ to a negative number, and so cannot be a positive linear functional. Thus the dual basis of a positive basis can only itself be positive if $V$ is classical.

So what does this mean? Just that $\hat{V}=\hom_{\mathcal{O}rd\mathcal{L}in}(V,\mathbb{F})$ is only the categorical dual to $V$ when $V$ is classical. Thus only in this case do we have an adjunction $V\otimes\underline{\hphantom{X}}\dashv\hat{V}\otimes\underline{\hphantom{X}}$, and so only in this case do we find $\hat{V}\otimes W\cong W^V$ naturally in $W$.

What happens for non-classical cones? Well, $(\underline{\hphantom{X}})^V$ is a right adjoint, and so its evaluation transformation $\epsilon_W:V\otimes W^V\rightarrow W$ is couniversal. Since we also have a natural transformation $V\otimes\hat{V}\otimes W\rightarrow W$ (evaluate the second factor on the first) it must factor through the couniversal. That is, there is an arrow $\hat{V}\otimes W\rightarrow W^V$, which is the one we earlier calculated explicitly. But it’s not generally an isomorphism.

September 25, 2007 Posted by | Category theory | 1 Comment

## New Blath!

Well, as today’s main post sort of died, I really should post some good news. Jeff Morton has started a weblog! It’s a great coincidence, because just yesterday I was mentioning his use of spans in categorifying the quantum harmonic oscillator during my talk. I see good things coming from this.

September 25, 2007 Posted by | Uncategorized | 2 Comments