# The forgetful functor $U: \text{Con}(\mathcal C) \to \mathcal C$ has a left adjoint.

Let $$\mathcal C$$ a category. We define the category of cones $$\text{Con}(\mathcal C)$$ in the following way:

• Objects: quadruples $$(\mathcal Z, \underline{M}, M, m)$$ consisting of a small category $$\mathcal Z$$, a functor $$\underline{M}: \mathcal Z \to \mathcal C$$ and a cone $$(M, m)$$ of $$\underline{M}$$ (i.e. $$M$$ is an object in $$\mathcal C$$ and $$m = (m_Z)_{Z \in \mathcal Z}$$ is a collection of morphism $$m_Z: M \to \underline{M}(Z)$$ satisfying a compatibility condition) .
• Morphisms: a morphism $$\underline{\phi}: (\mathcal Z, \underline{M}, M, m) \to (\mathcal Y, \underline{N}, N, n)$$ is a triple $$\underline{\phi}:( \phi^0, \phi^1, \phi^2)$$ where $$\phi^0: \mathcal Y \to \mathcal Z$$ is a functor, $$\phi^1: M \to N$$ is a morphism in $$\mathcal C$$ and $$\phi^2: \underline{M}\circ \phi^0 \to \underline{N}$$ is a natural transformation such that $$\phi^2_Y \circ m_{\phi^0(Y)} = n_Y \circ \phi^1$$ for all $$Y \in \mathcal Y$$.

I am trying to show that the forgetful functor $$U: \text{Con}(\mathcal C) \to \mathcal C$$ has a left adjoint, i.e. there is a functor $$S : \mathcal C \to \text{Con}(\mathcal C)$$ such that $$\text{Hom}_{\text{Con}(\mathcal C)}(S(X), Y) \cong \text{Hom}_\mathcal C (X, U(Y)).$$ The more natural way to associate an element $$C \in \mathcal C$$ to a quadruple in $$\text{Con}(\mathcal C)$$ is to send $$C$$ to $$(\varnothing, F_\varnothing, C, m)$$ where $$F_\varnothing$$ is the unique functor that goes from $$\varnothing$$ to $$\mathcal C$$ and $$(C, m)$$ is a cone for $$F_\varnothing$$ (a cone for this functor is just an element, there is no need to consider such an $$m$$).

This seems to be a natural way to define the functor $$S:\mathcal C \to \text{Con}(\mathcal C)$$ as I did above but I do not really see how it goes for the morphism. If we consider a morphism $$\underline{f} \in \text{Hom}_{\text{Con}(\mathcal C)}(S(C), Y)$$ (for $$S(C)$$ construct above and $$Y = (\mathcal Y, \underline{N}, N, n)$$ a quadruple in $$\text{Con}(\mathcal C)$$), $$\underline{f}$$ should be a triple $$\underline{f} = (f^0, f^1, f^2)$$ with $$f^0: \mathcal Y \to \varnothing$$ and $$f^3: F_\varnothing \circ f^0 \to \underline{N}$$. The problem is that $$F_\varnothing \circ f^0 : \mathcal Y \to \mathcal C$$ factors trough the empty set and therefore it has no image, $$F_\varnothing \circ f^0$$ is not even a functor.

My questions are the following: Is it really possible to consider a functor as $$f^0$$ such that $$f^0: \mathcal Y \to \varnothing$$ ? Is it the right way to construct the left adjoint $$S$$ ? I also tried to construct $$S$$ by sending $$C$$ to an other quadruble where the associate $$\mathcal Z$$ is not the empty set but it seems that in those cases there are many possiblities which are all very arbitrary (there is a unique functor $$\varnothing \to \mathcal C$$ but there are a lots a functor from $$\mathcal Z = \underline{1} \to \mathcal C$$, where $$\underline{1}$$ is the category with a unique element, for example).

• Can you clarify what "$(M,m)$ is a cone of $\underline{M}$" means for you? My best guess is that $M$ is an object in $\mathcal{C}$ and $m$ is a collection of morphisms from $M$ to the objects in the diagram $\underline{M}$ (satisfying the correct compatibility conditions) -- is that right? – diracdeltafunk Jan 15 at 19:59
• Yes that's it, is there a more "universal" notation for a cone of a functor ? It is the one we use in category theory course. – Falcon Jan 15 at 20:18
• I've edited, hope it is more clear now. – Falcon Jan 15 at 20:25
• Why do you define the category of cones like this? It would seem more natural to have the functor go from Z to Y instead (and the rest changed accordingly). – Thorgott Jan 15 at 23:49
• Since a morphism $\underline{\phi}: (\mathcal Z, \underline{M}, M, m) \to (\mathcal Y, \underline{N}, N, n)$ involves a functor $\mathcal{Y}\to\mathcal{Z}$, rather than the other direction, and since the left adjoint property of $S(C)$ involves morphisms from $S(C)$, and therefore functors to the category $\mathcal{Z}$ associated with $S(C)$, it makes sense to choose $\mathcal{Z}=\underline{1}$, since that has a unique functor to it from every other category. – Jeremy Rickard Jan 16 at 9:58

## 2 Answers

Your idea to take the cone over an empty category to represent objects is basically good, however, as was noted in the comments, a morphism from a cone over shape $$\mathcal Z$$ to a cone over $$\mathcal Y$$ involves a contravariant part, namely the functor $$\phi^0:\mathcal Y\to\mathcal Z$$.
This is because the hom functor (of the category of categories) is contravariant in the first argument.

So, instead of the initial category we need to consider the terminal category: the discrete category $$\bf 1$$ with one object $$\ast$$.
And, then for an object $$X\in\mathcal C$$, define $$S(X):=({\bf 1},\ \ast\mapsto X,\ X,\ 1_X)$$, and verify the claim.

I've tried to show the statement for $$S(C) = (\underline{1}, F_{C}, C, 1_C)$$ (where $$F_C$$ is the functor $$\{*\} \to C$$) as recommended by Berci and I wanted to know everything is good in the justification, let me know if you guys see anything wrong.

For $$\widetilde{Y} = (\mathcal Y, \underline{N}, N, n) \in \text{Con}(\mathcal C)$$, we can consider $$\underline{\phi} \in \text{Hom}_{\text{Con}(\mathcal C)}(S(C), \widetilde{Y})$$ so that $$\underline{\phi} = (\phi^0, \phi^1, \phi^2)$$ where $$\phi^0$$ is the unique functor that goes from $$\mathcal Y$$ to $$\underline{1}$$, $$\phi^1: C \to N$$ is any morphism in $$\text{Hom}_\mathcal C(C, N) = \text{Hom}_\mathcal C(C, U(\widetilde{Y}))$$ and $$\phi^2$$ is a natural transformation $$F_C \circ \phi^0 \to \underline{N}$$.

The natural transformation $$\phi^2$$ requires
$$\phi^2_Y = n_Y \circ \phi^1,$$ since $$m_{\phi^0(Y)} = 1_C$$. Therefore, $$\underline{\phi} \in \text{Hom}_{\text{Con}(\mathcal C)}(S(C), \widetilde{Y})$$ is entirely determined by a morphism $$\phi^1 \in \text{Hom}_\mathcal C(C, U(\widetilde{Y}))$$ since $$\phi^0$$ is unique and $$\phi^2_Y$$ is equal to $$n_Y \circ \phi^1$$. We conclude that $$\text{Hom}_{\text{Con}(\mathcal C)}(S(C), \widetilde{Y})\cong \text{Hom}_\mathcal C(C, U(\widetilde{Y})).$$ (The naturality in $$C, \widetilde{Y}$$ seems quite ugly so I'm gonna stop here).

• It's correct. To handle naturality cleanly, there's an approach for adjunctions via profunctors: specifically here we can add so called heteromorphisms $C\to(\mathcal Z,\underline M,M,m)$ to the disjoint union $\mathcal C+{\rm Con}(\mathcal C)$ simply as arrows $C\to M$ (define compositions so that we receive a category), then show that every object in $\mathcal C$ has the reflection $S(C)$ in the subcategory ${\rm Con}(\mathcal C)$ and every cone has its summit as its coreflection in the subcategory $\mathcal C$. – Berci Jan 16 at 22:24