5
$\begingroup$

In an article I am writing, I am led to the following generalization of the notion of functor. Let $C$ and $D$ and be two categories. A generalized functor $F : C \to D$ is given by:

  • a function $f : C_0 \to D_0$
  • for all $c \in C_0$, a set $F_c$
  • for all $c_1,c_2 \in C_0$, a function $F_{c_1,c_2}:C(c_1,c_2) \times F_{c_2} \to F_{c_1} \times D(f(c_1),f(c_2))$.

All this must be compatible with the product and units, that is:

  • for all $c \in C$ and all $x \in F_c$, $F_{c,c} (1_c, x) = (x,1_{f(c)})$
  • for all $c_1,c_2,c_3 \in C$, the following composites are equal:

$$C(c_1,c_2) \times C(c_2,c_3) \times F_{c_3} \to C(c_1,c_3) \times F_{c_3} \to F_{c_1} \times D(f(c_1),f(c_3)) $$ and $$C(c_1,c_2) \times C(c_2,c_3) \times F_{c_3} \to C(c_1,c_2) \times F_{c_2} \times D(f(c_2),f(c_3)) \to F_{c_1}\times D(f(c_1),f(c_2))\times D(f(c_2),f(c_3)) \to F_{c_1}\times D(f(c_1),f(c_3))$$

In particular, a functor is a generalized functor, where every $F_c$ is the singleton set.

Are you aware of a similar structure arising somewhere?

One way to see this structure is the following. One can see objects $a,b$ of a category as points, and then see $C(a,b)$ as an arrow form $a$ to $b$. If you say that the composition of two arrows is given by the Cartesian product, then you can see the composition in $C$ as a $2$-arrow from $C(a,b) \times C(b,c) \to C(a,c)$.

composition as a $2$-cell

Now how do you represent a functor in this setting? You can't represent the function $C(a,b) \to D(f(a),f(b))$ by a $2$-arrow, because the 'arrows' $C(a,b)$ and $D(f(a),f(b))$ are not parallel (the first one goes from $a$ to $b$, while the second goes from $f(a)$ to $f(b)$).

functor as an application

One way to deal with that is to add two arrows: one called $F_{a}$ from $a$ to $f(a)$ and the other called $F_{b}$ from $b$ to $f(b)$. Now we have a square that we can fill with a $2$-arrow $F_{a,b}$.

functor as a 2-cell

If we keep in mind the correspondence 'arrows are sets' and '$2$-arrows are functions' from the previous paragraph, we obtain the definition of a generalized functor.

Two more things to note:

  • One can define generalized natural transformations between generalised functors. The usual naturality arrows $\eta_a : f(a) \to g(a)$ are then replaced by functions $G_a \to F_a \times D(f(a),g(a))$:

natural transformation as a 2-cell

  • This can be extended to $n$-functors by replacing the sets $F_a$ by $(n-1)$-categories and the functions $F_{a,b}$ by $(n-1)$-functors.
$\endgroup$
  • 4
    $\begingroup$ Can you say intuitively what this structure does? $\endgroup$ – Jonathan Beardsley Mar 12 '15 at 13:20
  • $\begingroup$ @JonBeardsley I edited my question. I hope it is more clear. $\endgroup$ – Maxime Lucas Mar 12 '15 at 14:02
  • $\begingroup$ Do you have an example of such a thing that you care about? $\endgroup$ – Qiaochu Yuan Mar 13 '15 at 9:56
  • $\begingroup$ It's really hard to read if you write this in terms of hom-sets. How about drawing some diagrams to say what's going on? $\endgroup$ – Paul Taylor Mar 14 '15 at 10:36
  • $\begingroup$ @PaulTaylor I would love to, but I cannot find how to make diagrams work on MathOverflow. How do I do that? $\endgroup$ – Maxime Lucas Mar 17 '15 at 10:59
2
$\begingroup$

Here is one way to look at it: if V is a monoidal category and $\mathbf{B} V$ is the corresponding one-object bicategory, then a V-category in the usual sense is the same thing as a lax functor $\mathrm{c}(C_0) \to \mathbf{B} V$, where $C_0$ is a set and $\mathrm{c}(\cdot)$ is the fully faithful functor that takes a set to the codiscrete category on it (i.e. the one with $C_0$ as object set and exactly one morphism between each pair of objects). If C and D are V-categories then a V-functor from C to D is given by a function $f \colon C_0 \to D_0$ together with an identity-component oplax natural transformation (an icon) $\hat C \to \hat D \circ \mathrm{c}(f)$, where $\hat C, \hat D$ are the lax functors corresponding to C and D. What you have seems to be a general oplax transformation, without the condition that its components be identities. For that reason it's probably a not-unreasonable definition to write down, although I for one have never seen an example in the wild.

At the end of your question you are getting somewhere close to the notion of a category enriched in a bicategory, see here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.