Category of cones: what if morphisms to match natural transformations are missing?

Question

Suppose I have $\mathcal{J} = \mathbf{2}$, which is the discrete category with two objects. Now I have a category $\mathcal{C}$, with:

$$ c, c', c''\\ f: c \to c' \\ g: c \to c'' $$

Now I have a functor $D: \mathbf{1} \to \mathcal{C}^\mathcal{J}$, where $\mathcal{C}^\mathcal{J}$ is the category of diagrams. Additionally, I have a functor $\Delta: \mathcal{C} \to \mathcal{C}^\mathcal{J}$. From this, I want to construct a category of cones $\Delta \downarrow D$. In this category there are the following objects and morphisms:

$$ (c,\ast,h_c:\Delta c \to D\ast) \\ (c',\ast,h_{c'}:\Delta c' \to D\ast) \\ (c'',\ast,h_{c''}:\Delta c'' \to D\ast) \\ (f,1_\ast) \\ (g,1_\ast) $$ Now I want to concentrate on $h_c, h_{c'}, h_{c''}$. They all are morphisms in a functor category, which makes them natural transformations, or, as seen in context, cones. However looking back on $\mathcal{C}$, it turns out that not all of these are complete. E.g. $h_{c'}$ could have $1_{c'}$ as one of its natural transformations, but there is no way to get from $c'$ to $c''$ in $\mathcal{C}$, which makes the cone exist in $\mathcal{C}^\mathcal{J}$, but not in $\mathcal{C}$.

Is this okay? Where did I go wrong? Do I have to leave objects which don't form cones out of the category of cones?

Answer 1

The objects of $\Delta \downarrow D$ are all triples $(a,b,h)$ where $a$ is an object of $\mathcal{C}$, $b$ is an object of $\mathbf{1}$, and $h$ is a natural transformation $\Delta(a)\to D(b)$. In particular, such an object depends on a choice of $h$, not just on $a$ and $b$. For any particular choice of $a$ and $b$, there may be no natural transformation $h:\Delta(a)\to D(b)$, or there may be more than one. There is a separate object of $\Delta \downarrow D$ for each such choice.

So it doesn't make sense to refer to $(c,*,h_c)$ as an object until you have chosen a specific definition of $h_c$. If there is not any possible such definition, then there is no such object. If there is more than one such definition, then each one gives you a different object.

Answer 2

As explained in Eric's answer too, you have to check that the natural tranformations ( cones ) exist, before you build the objects in the comma category. For example, with your $C$ category consisting of 3 objects and 2 morphisms, $h_{c'}$ and $h_{c''}$ do not exist because there are no morphisms to form the cones (unless your $D$ is a constant functor ). In short your comma category has just one object (the first one you wrote).