Suppose you have two categories $C$ and $D$, and a functor $F: C \to D$. Then we can build a new category as follows:

First, take the union (or coproduct, if you like) of the categories $C$ and $D$.

Then, for each object $X$ in $C$, mapping to $F(X)$ in $D$, add a new morphism from $X \to F(X)$, and extend via composition, so that all the squares formed by composing the new morphisms with the old ones commute.

A good diagram that to make the last requirement clear is this picture from the nCatlab's page on "Functor":

functor picture

This is supposed to be their graphical depiction of a functor, but it is also kind of a good picture of the category I am talking about. Treat the entire picture as one big category, and then the dotted arrows are the new morphisms, so that all the squares thus generated commute (e.g. the square at $X, Y, F(X),$ and $F(Y)$).

Is there a name for this category? I have done some searching but have not seen a name for this. The best I've thought of so far is that perhaps it could be a very strange version of a slice category in some strange way.

  • $\begingroup$ You are surely starting with the coproduct of $C$ and $D$, not their product. $\endgroup$ Jul 12, 2020 at 5:05
  • $\begingroup$ Thanks - would it be the coproduct? I haven't seen the coproduct of a category, just of elements in a category. I edited it to say "union" $\endgroup$ Jul 12, 2020 at 5:11
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    $\begingroup$ The coproduct of categories is their "disjoint union" (just like with sets). $\endgroup$ Jul 12, 2020 at 5:13
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    $\begingroup$ This sounds like some kind of lax colimit of the diagram $C \stackrel{F}{\to} D$. Something similar to the mapping cylinder in homotopy theory. $\endgroup$
    – Zhen Lin
    Jul 12, 2020 at 6:00
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    $\begingroup$ If you need to invent a term, I like "functoring cylinder". Your diagram evokes this, and it is consistent with @ZhenLin 's analogy. $\endgroup$
    – tkf
    Jul 12, 2020 at 7:07

3 Answers 3


It sounds like you're looking for the collage category of the profunctor $F_*:C\not\to D,\ (c,d)\mapsto \hom_D(Fc,d)$ induced by the functor $F$:

There we freely add arrows $c\to Fc$ subject to making the following squares commutative for each arrow $u:c\to c'$: $$\matrix{c &\overset u\to& c'\\ \downarrow &&\downarrow \\ Fc& \underset{Fu}\to &Fc'}$$


Consider the category of categories $E$ equipped with functors: \begin{eqnarray*}G_E\colon C\to E,\\H_E\colon D\to E,\end{eqnarray*} and a natural transformation $\eta_E\colon G_E\to H_E\circ F$.

Let morphisms in this category be functors $K\colon E_1\to E_2$ satisfying:

Then your category along with natural inclusions $C\hookrightarrow E,\,D\hookrightarrow E$ is the initial object in this category.

  • $\begingroup$ This should probably be a comment rather than an answer but it got long. $\endgroup$
    – tkf
    Jul 12, 2020 at 6:22
  • $\begingroup$ Well, one can interpret this as a category of weighted cocones, and then this is exactly the weighted colimit description, so it’s certainly useful. $\endgroup$ Jul 12, 2020 at 20:09

This category is also known as a co-comma, which is a type of pushout of the span $D\leftarrow C\to C$. Concretely, it given by taking the weighted colimit with weight $\bullet \to (\bullet \to \bullet)\leftarrow \bullet$, where the first inclusion is at the terminal and the second at the initial object. This gives your construction’s universal property: it represents triples $(f,g,\alpha)$ of a functor $f:C\to E,$ a functor $g:D\to E$, and a natural transformation$ \alpha:f \Rightarrow g\circ F$. It is not precisely a lax pushout, although they’re closely related.

I don’t know of a great literature reference for co-commas among categories, but there are a couple of relevant questions here and on MO:


How Co-Comma Categories are constructed?


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