Let $F : C \to E$ and $G : D \to E$ be functors. Consider the comma category $(F \downarrow G)$ with its projections $\pi_1 : (F \downarrow G) \to C$ and $\pi_2 : (F \downarrow G) \to D$. Using the elementary construction of a comma category, it is easy to prove the following equality: $$F \; \circ \; \pi_1 \;\; = \;\; G \; o \; \pi_2$$

However, a comma category can also be seen as a special case of weighted limit in the 2-category of 1-categories. How to prove the above equality by using the properties of a weighted limit? What else can we deduce about comma categories from the fact that they are weighted limits?


The equality you state is not true in general. An object $(c,f,d)$ of the comma category consists of an object $c$ of $C$, an object $d$ of $D$, and a morphism $f \colon Fc \longrightarrow Gd$ in $E$. The projection functors $\pi_1$ and $\pi_2$ send this object to $\pi_1(c,f,d) = c$ and $\pi_2(c,f,d) = d$, so we have $(F\circ\pi_1)(c,f,d) = Fc$ and $(G\circ\pi_2)(c,f,d) = Gd$, which are not equal in general. They will be equal if $E$ is discrete, in which case the comma category is the pullback.

What is true in general is that there is a natural transformation $\lambda \colon F\circ\pi_1 \longrightarrow G\circ\pi_2$. The component of $\lambda$ at the object $(c,f,d)$ of $F \downarrow G$ is the morphism $f \colon Fc \longrightarrow Gd$ in $E$. This natural transformation is part of the structure of the comma category as a weighted limit, as I will now spell out.

The limit of a $2$-functor $J \colon \mathcal{A} \longrightarrow \mathcal{B}$ weighted by a $2$-functor $W \colon \mathcal{A} \longrightarrow \mathbf{\text{Cat}}$ consists of an object $\{W,J\}$ of $\mathcal{B}$ together with a ''universal cylinder'', which is a $2$-natural transformation $W \longrightarrow \mathcal{B}(\{W,J\},J-)$ such that the induced composite functor $$ \mathcal{B}(B,\{W,J\}) \longrightarrow [\mathcal{A},\mathbf{\text{Cat}}](\mathcal{B}(\{W,J\},J-),\mathcal{B}(B,J-)) \longrightarrow [\mathcal{A},\mathbf{\text{Cat}}](W,\mathcal{B}(B,J-))$$ is an isomorphism of categories.

(Note that if $\mathcal{A}$ is a category, then these $2$-functors and $2$-natural transformations just amount to functors and natural transformations respectively.)

Now let $\mathcal{A}$ denote the free category on the graph $\bullet \longrightarrow \bullet \longleftarrow \bullet$.The comma object $f/g$ of a cospan $f \colon A \longrightarrow C \longleftarrow B \colon g$ in a $2$-category $\mathcal{B}$ is the limit of the corresponding functor $J \colon \mathcal{A} \longrightarrow \mathcal{B}$ weighted by the functor $W \colon\mathcal{A} \longrightarrow \mathbf{\text{Cat}}$ that picks out the cospan $\bot \colon 1 \longrightarrow \mathbf{2} \longleftarrow 1 \colon \top$, where $\mathbf{2}$ denotes the free category on the graph $\bot \longrightarrow \top$. The universal cylinder for this weighted limit is a natural transformation as displayed in the commutative diagram $$ \require{AMScd} \begin{CD} 1 @>\bot>> \mathbf{2} @<\top<< 1\\ @VpVV @VV\lambda V @VVqV\\ \mathcal{B}(f/g,A) @>>\mathcal{B}(1,f)> \mathcal{B}(f/g,C) @<<\mathcal{B}(1,g)< \mathcal{B}(f/g,B) \end{CD}$$ in $\mathbf{\text{Cat}}$, which amounts to a $2$-cell $\lambda \colon fp \longrightarrow gq$ in $\mathcal{B}$ as in the following diagram. $$ \require{AMScd} \begin{CD} f/g @>q>> B \\ @VpVV \Longrightarrow @VVgV \\ A @>>f> C \end{CD} $$

  • $\begingroup$ I understand that the natural transformation $\lambda \colon F\circ\pi_1 \longrightarrow G\circ\pi_2$ comes from the 2-naturality of the universal cylinder. What about the universal property (in its 1-dimensional and 2-dimensional aspects) of the weighted limit? What does it say about the comma category? $\endgroup$ – Bob Oct 25 '16 at 18:14
  • 1
    $\begingroup$ Note that since our $\mathcal{A}$ is a category, we can just think of the universal cylinder as an ordinary natural transformation, there is no $2$-dimensional condition on it at all. You can read off the universal property of the universal cylinder from the third paragraph of my answer. It is that for each object $X$ of the $2$-category $\mathcal{B}$, the universal cylinder induces an isomorphism between the hom-category $\mathcal{B}(X,f/g)$ and the category of ''cylinders with vertex $X$'', which are squares like the final diagram, but with $X$ in place of $f/g$. $\endgroup$ – Alexander Campbell Oct 25 '16 at 22:25

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.