Absolute Colimits

May 18, 2019 23:59 UTC (Last updated on August 2, 2024 02:42 UTC)

Tags: math, category theory

In category theory a concept is called absolute if it is preserved by all functors. Identity arrows and composition are absolute by the definition of functor. Less trivially, isomorphisms are absolute. In general, anything that is described by a diagram commuting is absolute as that diagram will be preserved by any functor. This is generally the case, but if I tell you something is an absolute epimorphism, it’s not clear what diagram is causing that; the notion of epimorphism itself doesn’t reduce to the commutativity of a particular diagram.

Below I’ll be focused primarily on absolute colimits as those are the most commonly used examples. They are an important part of the theory of monadicity. The trick to many questions about absolute colimits and related concepts is to see what it means for $\mathsf{Hom}$ functors to preserve them.

Non-Examples

To start, we can show that certain colimits cannot be absolute, at least for $\mathbf{Set}$-enriched category theory. In particular, initial objects and coproducts are never absolute. Using the trick above, this is easily proven.

$$\mathsf{Hom}(0,0)\cong 1\ncong 0$$

$$\mathbf{Set}(\mathsf{Hom}(X+Y,Z),1)\cong 1\ncong 2\cong \mathbf{Set}(\mathsf{Hom}(X,Z),1)+\mathbf{Set}(\mathsf{Hom}(Y,Z),1)$$

Absolute Epimorphisms

What do absolute epimorphisms look like? We know that there are absolute epimorphisms because a split epimorphism is defined by a certain diagram commuting. Are there other absolute epimorphisms? To find out, we apply our trick.

Let $r:X\twoheadrightarrow Y$ be an arbitrary absolute epimorphism. Absoluteness implies the surjection $$\mathsf{Hom}(Y,r):\mathsf{Hom}(Y,X)\twoheadrightarrow \mathsf{Hom}(Y,Y)$$ but this means that for every arrow $f:Y\to Y$, there’s an arrow $s:Y\to X$ such that $f=r\circ s$. As you can no doubt guess, we want to choose $f=i{d}_Y$, and we then have that $r$ is a split epimorphism. Therefore split epimorphisms are the only examples of absolute epimorphisms.

Split Coequalizers

Now let’s consider the coequalizer case. Let $f,g:X\to Y$ and $e:Y\twoheadrightarrow C$ be their coequalizer which we’ll assume is absolute. Before we pull out our trick, we can immediately use the previous result to show that $e$ has a section, i.e. an arrow $s:C\rightarrowtail Y$ such that $i{d}_C=e\circ s$. Moving on, we use the trick to get the diagram: $$\mathsf{Hom}(Y,X)\rightrightarrows \mathsf{Hom}(Y,Y)\twoheadrightarrow \mathsf{Hom}(Y,C)$$

Next, we use the explicit construction of the coequalizer in $$ which $$ is supposed to be canonically isomorphic to. That is, the coequalizer of $$ and $$ is $$ quotiented by the equivalence relation generated by the relation, $$, which identifies $$ when $$. Let $$ represent the equivalence class of $$. That is, $$ corresponds to the function $$, where $$ is the reflexive, symmetric, transitive closure of $$, witnessing the set-theoretic coequalizer. The claim that $$ is (with $$) a coequalizer of the above arrows implies that we have a canonical isomorphism $$ such that $$ and thus $$. Of course, our next move is to choose $$ giving $$. However, $$ so we get $$ because $$ is invertible.

Expanding the definition of $$, we have $$. Therefore $$. Now let’s make a simplifying assumption and assume further that $$, i.e. that $$ is directly related to $$ by $$. By definition of $$ this means there is a $$ such that $$ and $$. A given triple of $$, $$, and $$ such that $$ and equipped with a $$ and $$ satisfying the previous two equations along with $$ is called a split coequalizer. This data is specified diagrammatically and so is preserved by all functors, thus split coequalizers are absolute. All that we need to show is that this data is enough, on its own, to show that $$ is a coequalizer.

Given any $$ such that $$, we need to show that there exists a unique arrow $$ which $$ factors through via $$, i.e. $$ for some (unique) $$. The obvious candidate is $$ leading to us needing to verify that $$. We calculate as follows: $$ Uniqueness then quickly follows since if $$ then $$. $$

There’s actually an even simpler example where $$ which corresponds to the trivial case where $$.

Absolute Coequalizers

Split coequalizers show that (non-trivial) absolute coequalizers can exist, but they don’t exhaust all the possibilities. The obvious cause of this is the simplifying assumption we made where we said $$ rather than $$. In the general case, the equivalence will be witnessed by a sequence of arrows $$ such that we have either $$ or $$, then $$ or $$ respectively, and so on until we get to $$ or $$. As a diagram, this is a fan of diamonds of the form $$ or $$ with a diamond with side $$ on one end of the fan and a triangle with $$ on the other. All this data is diagrammatic so it is preserved by all functors making the coequalizer absolute. That it is a coequalizer uses the same proof as for split coequalizers except that we have a series of equalities to show that $$, namely all those pasted together diamonds. There is no conceptual difficulty here; it’s just awkward to notate.

Absolute Colimits

The absolute coequalizer case captures the spirit of the general case, but you can see an explicit description of the general case here. I’m not going to work through it, but you could as an exercise. Less tediously, you could work through absolute pushouts. If $$ is the pushout of $$, then the functors to consider are $$ and $$. For each, the pushout in $$ can be turned into a coequalizer of a coproduct. For the first functor, as before, this gives us an inverse image of $$ which will either be an arrow $$ or an arrow $$ which will play the role of $$. The other functor produces a coequalizer of $$. The generating relation of the equivalence relation states that there exists either an arrow $$ or an arrow $$ such that the appropriate equation holds. The story plays out much as before except that we have a sequence of arrows from two different hom-sets.