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).
 A: 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.
A: 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).
