Many categories can be defined as weighted limits or colimits in the 2-category of categories Cat. For example the category 1 (one object with its identity) is the terminal object of Cat, the category 2 (two object with their respective identities) is the coproduct 1+1. Even comma categories are weighted limits in Cat.

What about the category {0 --> 1} consisting of two objects 0 and 1 with their identities and one morphism between them? Can it be defined as some kind of limit (or composition of limits like 1+1) in Cat?

  • $\begingroup$ A similar question arose regarding one of my answers. Maybe Zhen Lin's following comment can be made into a construction of limits and colimits: math.stackexchange.com/questions/782426/… $\endgroup$
    – Berci
    Nov 1, 2016 at 22:14

1 Answer 1


First, a trivial remark. Any object $c$ of a category $C$ is both the colimit and the limit of the functor $c \colon 1 \longrightarrow C$ that sends the unique object of the terminal category $1$ to $c$. So in particular, the category $\mathbf{2} = (0\to 1)$ is both the colimit and the limit of the functor $\mathbf{2} \colon 1 \longrightarrow \mathbf{\text{Cat}}$.

Now, more interesting, as you suggest in your question, is when we can build up an object as a colimit of simpler objects. We can see the category $\mathbf{2}$ as a weighted colimit in $\mathbf{Cat}$ in a number of ways, for example:

  1. as the tensor $\mathbf{2}\ast 1$, i.e. the colimit of the functor $1 \colon 1 \longrightarrow \mathbf{\text{Cat}}$ weighted by the functor $\mathbf{2} \colon 1 \longrightarrow \mathbf{\text{Cat}}$;

  2. as the coinserter of the parallel pair $0,1 \colon 1 \longrightarrow 2$ (here $2$ is the discrete category with objects $0$ and $1$), which is the colimit weighted by the parallel pair $0,1\colon 1 \longrightarrow \mathbf{2}$;

  3. as the co-comma category $$ \require{AMScd} \begin{CD} 1 @>1>> 1 \\ @V1VV \Longrightarrow @VV1V \\ 1 @>>0> \mathbf{2} \end{CD} $$ i.e. the colimit of the span $1 \longleftarrow 1 \longrightarrow 1$ weighted by the cospan $0 \colon 1 \longrightarrow \mathbf{2} \longleftarrow 1 \colon 1$. In fact this follows from the previous example, since co-comma categories can in general be constructed by first taking a coproduct ($1+1 \cong 2$) and then taking a coinserter (as above);

  4. as the colimit of the arrow $1 \longrightarrow 1$ (seen as a functor $\mathbf{2} \longrightarrow \mathbf{\text{Cat}}$) weighted by the arrow $1 \colon 1 \longrightarrow \mathbf{2}$. (For reasons to be hinted at below, this kind of weighted colimit is called a "lax colimit of an arrow".)

Hence the category $\mathbf{2}$ can be seen in a number of ways as a weighted colimit of discrete category (i.e. set)-valued diagrams, and moreover of diagrams with constant value $1$. However this is in some sense not entirely satisfying, since $\mathbf{2}$ already features in the weights of all these colimits. We can remedy this concern by considering lax colimits in $\mathbf{\text{Cat}}$.

The lax colimit of a functor $F \colon A \longrightarrow \mathbf{\text{Cat}}$ is a category $\texttt{laxcolim}F$ together with a lax natural transformation $F \longrightarrow \Delta (\texttt{laxcolim}F)$, such that for each category $C$, the induced functor $$[\texttt{laxcolim}F,C] \longrightarrow \text{Lax}[A,\mathbf{\textbf{Cat}}](F,\Delta C)$$ is an isomorphism. Here the domain is the category of functors $\texttt{laxcolim}F \longrightarrow C$ and natural transformations between them, and the codomain is the category of lax natural transformations $F \longrightarrow \Delta C$ and modifications between them.

By spelling out the definition, one can see that any category $A$ is the lax colimit of the functor $\Delta 1 \colon A \longrightarrow \mathbf{\text{Cat}}$ with constant value $1$. In particular, the category $\mathbf{2}$ is the lax colimit of the constant functor $\Delta 1 \colon \mathbf{2}\longrightarrow \mathbf{\text{Cat}}$, also known as the lax colimit of the arrow $1 \longrightarrow 1$.

(Note that there is a general construction by which a lax colimit can be calculated as a weighted colimit; for our description of $\mathbf{2}$ as a lax colimit, this construction retrieves the weighted colimit of example 4.)

  • 3
    $\begingroup$ In all those definitions of 2 (even the lax-colimit one), 2 appears in its own definition! Does it mean that 2 can be defined as a least (or greatest) fixpoint? $\endgroup$
    – Bob
    Nov 4, 2016 at 7:48
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    $\begingroup$ @Bob The above colimits are not intended to be the "definition" of the category $\mathbf{2}$. The fundamental universal property of the category $\mathbf{2}$ is that it represents the functor $\mathbf{\text{Cat}} \longrightarrow \mathbf{\text{Set}}$ that sends a category to its set of morphisms; this suffices to define $\mathbf{2}$ up to isomorphism. But why not just define it is as the category with two objects and one non-identity morphism? $\endgroup$ Nov 4, 2016 at 8:40
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    $\begingroup$ Because I would like to define 2 without involving sets. A bit like it is possible to define an adjunction without involving sets: by using Lawvere's definition involving comma categories. $\endgroup$
    – Bob
    Nov 4, 2016 at 9:03
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    $\begingroup$ @Bob Also note that just as one can unpack the definition of the colimit of the functor $\Delta 1 \colon 2 \longrightarrow \mathbf{\text{Set}}$ to remove explicit reference to the set $2$, one can similarly unpack the definitions colimits in examples 2,3,4 and the lax colimit to remove explicit reference to the category $\mathbf{2}$. For just as we can say "the coproduct $1+1$", we can also say "the lax colimit of the arrow $1\longrightarrow 1$", as I indeed did in my answer. $\endgroup$ Nov 4, 2016 at 9:12
  • 1
    $\begingroup$ @Bob I strongly encourage you to unpack the universal property of the lax colimit of an arrow, you will see that this is an elementary and simple statement just using the 2-category structure of Cat. (You need only to concern yourself with the assertion that the displayed functor in the general definition of lax colimit above be an isomorphism on objects.) $\endgroup$ Nov 4, 2016 at 9:29

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