Explanation of terminal objects using limits

Question

I understand terminal objects, but I falter when trying to achieve them using the approach of diagrams, the category of constant diagrams $\mathcal{C}^{\mathcal{J}}$ and limits.

I have the empty category 0, now I have a diagram $D: \mathcal{J} \to \mathcal{C}$. However this diagrams image should be empty, because there are no objects or morphisms to map. At this point, I'm confused, because I can't get to the universal property $\forall c \in \mathcal{C}. \exists! f: c \to T$ anymore, because I'm missing T. (EDIT: I get that this is how it is supposed to be)

But continuing with the category of constant diagrams, even if I have a constant diagram mapping everything to one object / morphism, there is nothing to map, so the functors image is still empty. So now I'm missing both $T$, and any kind of $c$.

EDIT: It seems to me, have I chosen 1, instead of 0, I would have come up with that is considered the definition of the terminal category. What am I missing?

Answer 1

Careful: a limit of a diagram $D : \mathcal{J} \to \mathcal{C}$ is not necessarily an object in the image of $D$, it's simply an object of $\mathcal{C}$ (equipped with additional data).

Let's look carefully at the definition of a limit of $D : \mathcal{J} \to \mathcal{C}$.

First, it is a cone over $D$, meaning that it consists of:

• An object $T$ of $\mathcal{C}$;
• For each object $j$ of $\mathcal{J}$, a morphism $\mu_j : T \to D(j)$ in $\mathcal{C}$;

such that for each $\alpha : j \to j'$ in $\mathcal{J}$, we have $D(\alpha) \circ \mu_j = \mu_{j'}$.

When $\mathcal{J} = \mathbf{0}$, there are no objects $j$, and so there are no morphisms $\mu_j$, and so a cone is just an object of $\mathcal{C}$. Furthermore, there are no morphisms $\alpha : j \to j'$, and so the condition $D(\alpha) \circ \mu_j = \mu_{j'}$ for all $\alpha : j \to j'$ is satisfied vacuously.

What makes a cone $(T, \mu_j : T \to D(j))$ a limiting cone is: for all other cones $(C, \kappa_j : C \to D(j))$, there is a unique morphism $u : C \to T$ such that $\mu_j \circ u = \kappa_j$ for all $j$.

As we have seen, when $\mathcal{J} = \mathbf{0}$, a cone is just an object $C$, and the condition $\mu_j \circ u = \kappa_j$ holds vacuously. So the condition that $T$ be a limit for the empty diagram says precisely that, for all other objects $C$, there is a unique morphism $u : C \to T$.

...oh wait, that's exactly the definition of what it is for $T$ to be a terminal object!

Answer 2

A diagram $D$ from the empty category to $C$ consists of no arrows and no objects. For each object $c$ in $C$ there is precisely one cone from $c$ to $D$. It has no arrows. Each arrow $c\to c'$ in $C$ is a map between these empty cones. So limit of $D$ is the $c$ corresponding to a terminal object in this category of cones, that is a terminal object in $C$.

Answer 3

Consider the empty diagram $D$. Since $C$ (the codomain category of the diagram) has finite limits, this includes limits of empty diagrams. This means, by definition, that there exists an object $c \in \text{Ob}(C)$ such that for all morphisms from another object $x$ to the diagram $D$ (there are none, so it's a vacuous argument), there exists a unique morphism $u : x \to c$ such that the drawn triangles commute (but there are none), which is to say equivalently that there just exists a unique morphism $u: x \to c$. This can be said for all objects $x \in \text{Ob}(C)$ because each vacuously has an empty collection of morphisms from $x$ to the diagram $D$ (the image of $D$ in $C$ when drawn on paper). But that is precisely the definition of terminal object.

Therefore a terminal object is a limit of the empty diagram. Further, if $C$ has all finite limits, then $C$ must contain a terminal object (among other things such as pullbacks, etc).