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?


3 Answers 3


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!

  • $\begingroup$ Okay, I get the definition with cones, but I am still confused with the approach with the category of diagrams $\mathcal{C}^{\mathcal{J}}$, containing constant diagrams , I used the term "cone" in my description, which I think confused everyone. $\endgroup$
    – hgiesel
    Nov 30, 2017 at 20:31
  • $\begingroup$ When $\mathcal{J}=\mathbf{0}$, the category of diagrams is $\mathcal{C}^{\mathbf{0}}$, which is equivalent to the terminal category $\mathbf{1}$. You can check that a cone is precisely a functor $T : \mathbf{1} \to \mathcal{C}$, which is just an object of $\mathcal{C}$, and the condition that makes the cone a terminal cone is precisely the condition that the object picked out by $T$ is a terminal object (using arguments similar to the one given in my answer). $\endgroup$ Nov 30, 2017 at 20:35
  • $\begingroup$ Specifically, a limit $T$ is a right adjoint to the unique functor $\mathcal{C} \to \mathbf{1}$, so consists of a functor $T : \mathbf{1} \to \mathcal{C}$ (which we'll identify with an object $T$ of $\mathcal{C}$) and, for each object $C$ of $\mathcal{C}$, a morphism $\eta_C : C \to T$ such that, for all other morphisms $f : C \to T$, we have $f = \eta_C$... but that's exactly the assertion that for all $C$ there is a unique $C \to T$. $\endgroup$ Nov 30, 2017 at 20:39
  • $\begingroup$ But why does the category of diagrams contain diagrams $\mathbf{1} \to \mathcal{C}$? Shouldn't it contain functors $\mathcal{J} \to \mathcal{C}$, which would be $\mathbf{0} \to \mathcal{C}$ $\endgroup$
    – hgiesel
    Nov 30, 2017 at 20:39
  • $\begingroup$ @hgiesel: The category of diagrams $\mathcal{C}^{\mathcal{J}}$ has as its objects functors $\mathcal{J} \to \mathcal{C}$. However, a cone (and more generally a limit) is given by a functor $\mathcal{C}^{\mathcal{J}} \to \mathcal{C}$ (which is right adjoint to $\Delta : \mathcal{C} \to \mathcal{C}^{\mathcal{J}}$). When $\mathcal{J}=\mathbf{0}$, we have $\mathcal{C}^{\mathcal{J}} \simeq \mathbf{1}$ so that a limit is exactly a functor $\mathbf{1} \to \mathcal{C}$ satisfying certain conditions; and moreover, the 'diagonal functor' $\Delta : \mathcal{C} \to \mathbf{1}$ is uniquely determined. $\endgroup$ Nov 30, 2017 at 20:40

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 $C$.


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).


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