# Graphic intuition for generalizing to weighted limits

One of the ways to define a limit of a functor $F:\mathsf C\longrightarrow\mathsf D$ is a representation of $\mathsf{Nat}(\Delta-,F)$.

Along the journey of generalization to the enriched setting, one notes there's a bijection natural in $d$: $$\mathsf{Nat}(\Delta d,F)\cong \mathsf{Nat}(\Delta\mathbf 1,\mathsf D(d,F-)).$$

Thus a limit may be equivalently defined as representing of $\mathsf{Nat}(\Delta\mathbf 1,\mathsf D(-,F-)):\mathsf{D}^\text{op}\longrightarrow \mathsf{Set}$.

Now, the diagram encoded by $\mathsf{D}(d,F-)$ is just the canonical projection $d/F\rightarrow \mathsf C$, which is the base of a cone with vertex $d$ (the vertex is forgotten). Hence, I would think that a natural transformation $\Delta d\Rightarrow F$ should be the same as a natural transformation out of the "one object diagram" at $d$", i.e just $d$, into $\mathsf D(d,F-)$.

However, the diagram encoded by $\Delta\mathbf 1$ is $\mathsf C$ itself, so a natural transformation $\Delta\mathbf 1\Rightarrow\mathsf D(d,F-)$ doesn't graphically seem to give a cone.

What am I missing here?

• What do you mean by "the diagram encoded by"? – Kevin Carlson Jun 20 '16 at 21:20
• I don't understand the purpose of the question. You clearly know why such a thing is called a cone. Is it such a big problem that when you look at the same thing from a different perspective it ceases to look like a cone? – Zhen Lin Jun 20 '16 at 22:38
• @ZhenLin but it doesn't cease to look like a cone, the OP just made a mistake drawing the picture and is literally asking for help identifying the mistake. What perspective are you referring to from which (weighted) cones do not look like cones, at least over Set? – Vladimir Sotirov Jun 21 '16 at 3:30
• @KevinCarlson by "encoded diagram" I mean the term on page 8 of this document. – Arrow Jun 21 '16 at 14:55

## 1 Answer

Be explicit: categorical definitions tend to hide intuition as implementation details of the abstract structure described: it's very easy to get confused regarding what the definitions actually capture (as you do when it comes to interpreting the functor $D(d,F-)$). In particular, when using the correct interpretation of $D(d,F-)$, a natural transformation $\Delta\mathbf 1\Rightarrow D(d,F-)$ does indeed look like a cone.

Explicitly, $D(d,F-)$ takes $X\mapsto \{d\to FX\}$ and $X\xrightarrow{f} Y$ to $\{d\to FX\}\xrightarrow{Ff_*}\{d\to FY\}$; in particular this is certainly not the projection $d/F\to C$, but the fibers of the projection, so it's an "inverse functor" the same way that an inverse function $Y\to X$ gives preimages $f^{-1}(x)=\{y\in Y:f(y)=x\}$.

The relevance of the comma category $d/F$ (and of the "inverse functor" of its projection) to cones and weighted cones is this. The objects of $d/F$, i.e. morphisms $d\to FX$, are the exactly the projection morphisms that cones (and weighted cones) with vertex $d$ are made out of. A cone with vertex $d$ for $F$ is nothing other than a coherent choice of a single object of $d/F$ for each object $X$ (the morphism structure on $d/F$, and hence of $D(d,F-)$ is what allows us to say when such a choice is coherent).

In other words, to give a cone with vertex $d$ over $F$ means that for each object $X$ of $C$ we have to make a choice in $D(d,FX)=\{d\to FX\}$, and these choices have to be coherent. Such choices are made by giving morphisms $\mathbf 1\to D(d,FX)=\{d\to FX\}$ for each object $X$ in $C$, and the coherence of the choice is exactly the statement that these are the components of a natural transformation $\Delta\mathbf 1\Rightarrow D(d,F-)$.

Hence, natural transformations $\Delta\mathbf1\Rightarrow D(d,F-)$ are exactly cones with vertex $d$ over the diagram $F$, assembled by coherently choosing projection morphisms to $FX$ for each $X$ in $C$.

More generally, if you have a functor $C\xrightarrow{W}Set$, a natural transformation $W\Rightarrow D(d,F-)$ is exactly a weighted cone with vertex $d$, that is, a choice for each $X$ of a $W(X)$-(pointwise-)indexed family of morphisms $d\to FX$, coherent in the sense that post-composing with $FX\xrightarrow{Ff}FY$ takes an element indexed by $w_X\in W(X)$ to an element indexed by $Wf(w_X)$ in $W(Y)$. In other words, the weights of weighted cone are distributed on the projection morphisms of the cones.

• Thanks for the brilliant answer! What I meant by "encoded diagram" is the same as in page 8 of this document. In that sense, I think it is true that $\mathsf D(d,F-)$ encodes the projection diagram. Is this also a mistake? – Arrow Jun 21 '16 at 14:40
• The "encoded diagram" (ugh) of a functor $C^{op}\xrightarrow{F} Set$ is the unique diagram such that $F=colim C(-,X)$. The notes abuse $(C\downarrow F)$ to denote what should properly be denoted by $(Y_C\downarrow F)$ where $C\xrightarrow{Y_C}[C^{op},Set]$ is the (covariant) Yoneda embedding. Note that $(C\downarrow F)$ doesn't even make sense because the codomain of $F$ is wrong. I'm not sure off the top of my head exactly what happens if you switch variance and plug in $D(d,F-)$ in the place of $F$, but it seems unlikely you get the projection diagram you're thinking (work it out and see!). – Vladimir Sotirov Jun 21 '16 at 15:55
• I'm pretty sure the diagram "encoded" by a presheaf represented by $c$ is $\mathsf C/c$. This also makes sense to me because the Yoneda lemma can be formulated as $\mathsf C$ is dense in $\mathsf C$. For the opposite variance I think $\mathsf C(c,-)$ encodes $c/\mathsf C$ (seen as a subcategory of $\mathsf C$), and then that $\mathsf D(d,F-)$ encodes the projection I mentioned. This also makes sense because $\mathsf D(d,F-)$ - seen as an inverse image - involves the very same projection. Am I mistaken? – Arrow Jun 21 '16 at 16:04