If one views a group as a one object category with the elements of the group as morphisms then a natural transformation between functors of such categories is an inner automorphism, i.e. if we have two group homomorphisms $f,g: A\to B$ then a natural transformation $\eta :f\to g$ is just an element $b\in B$ such that $f(a)\cdot b = b \cdot g(a)$ which can be rewritten as $f(a)=b \cdot g(a)\cdot b^{-1}$. This isn't the only way to turn groups into categories. Another way is to take the elements of the group as objects and to have a morphism $h_a:a\to b$ if $h\cdot a=b$. If we view groups in this way then are the natural transformations again something nice like inner automorphisms?
[Math] groups as categories and their natural transformations
ct.category-theorygr.group-theory
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Theo, the answer is basically "yes". It's a qualified "yes", but only very lightly qualified.
Precisely: if a natural transformation between functors $\mathcal{C} \to \mathcal{D}$ is pointwise epi then it's epi. The converse doesn't always hold, but it does if $\mathcal{D}$ has pushouts. Dually, pointwise mono implies mono, and conversely if $\mathcal{D}$ has pullbacks.
The context for this --- and an answer to your more general question --- is the slogan
(Co)limits are computed pointwise.
You have, let's say, two functors $F, G: \mathcal{C} \to \mathcal{D}$, and you want to compute their product in the functor category $\mathcal{D}^\mathcal{C}$. Assuming that $\mathcal{D}$ has products, the product of $F$ and $G$ is computed in the simplest possible way, the 'pointwise' way: the value of the product $F \times G$ at an object $A \in \mathcal{C}$ is simply the product $F(A) \times G(A)$ in $\mathcal{D}$. The same goes for any other shape of limit or colimit.
For a statement of this, see for instance 5.1.5--5.1.8 of these notes. (It's probably in Categories for the Working Mathematician too.) See also sheet 9, question 1 at the page linked to. For the connection between monos and pullbacks (or epis and pushouts), see 4.1.31.
You do have to impose this condition that $\mathcal{D}$ has all (co)limits of the appropriate shape (pushouts in the case of your original question). Kelly came up with some example of an epi in $\mathcal{D}^\mathcal{C}$ that isn't pointwise epi; necessarily, his $\mathcal{D}$ doesn't have all pushouts.
I don't think I have seen anything like this published before, but I have written up a similar definition here (see here too).
One thing in your definition I would take issue with is that your 2-coend's universal property needs more to deserve the '2-' in its name: what I would expect (and what my definition, which otherwise seems to be equivalent to yours, says) is a bicategorical universal property that says
- for $(C', \mathrm{gl'}, \ldots)$ as above, $\eta$ exists and is invertible, and
- for $\theta, \theta' \colon C \to C'$, any 2-cell (i.e. modification between extranaturals/wedges) $\theta \circ \mathrm{gl} \to \theta' \circ \mathrm{gl}$ is equal to $t \circ \mathrm{gl}$ for a unique $t \colon \theta \to \theta'$.
Best Answer
The comments thread is getting a bit long, so here's an answer. The category $C(G)$ that David associates to a group $G$ (by his second recipe) has the elements of $G$ as its objects, and exactly one morphism between any given pair of objects. It's what category theorists call an indiscrete or codiscrete category, and graph theorists call a complete graph or clique. You can form the indiscrete category on any set: it doesn't need a group structure.
A functor from one indiscrete category to another is simply a function between their sets of underlying objects. In particular, given groups $G$ and $H$, a functor from $C(G)$ to $C(H)$ is simply a function from $G$ to $H$. That's any function (map of sets) whatsoever -- it completely ignores the group structure.
Given indiscrete categories $C$ and $D$ and functors $P, Q: C \to D$, there is always exactly one natural transformation from $P$ to $Q$. In particular, given groups $G$ and $H$ and functors $P, Q: C(G) \to C(H)$, there is always exactly one natural transformation from $P$ to $Q$.