If $T$ is terminal object, is also $\text{hom}(X,T)$ a terminal object?

Question

In a category with internal Hom and terminal object $T$. Is it true that also $\text{hom}(X,T)$ is a terminal object for any object $X$?

It is definitely true for $\mathbf{Set}$ or $\mathbf{Vec}$, but I'm not sure if it is true in general.

I'm probably stuck in thinking about sets, and I do not even know where to start in proving or disproving it.

Answer 1

I'm not really sure why none of the existing answers bothered to say it, but this is a special case of an extremely general and useful result. Namely, that right adjoints preserve limits.

If the internal hom is characterized by $\mathcal C(X\times Y,Z)\cong\mathcal C(X,[Y,Z])$ natural in $X$ and $Z$ (and usually also $Y$) or, more generally, $\mathcal C(X\otimes Y,Z)\cong\mathcal C(X,[Y,Z])$, this is just the statement that $X\times -$ (or $X\otimes-$) is left adjoint to $[Y,-]$. That is, $[Y,-]$ is a right adjoint. Right adjoints preserve all limits, and the terminal object is a limit. Preservation of limits is called continuity, so we have $[Y,1]\cong 1$ immediately by continuity. The proof of the general statement, i.e. that right adjoints are continuous, is very easy. Indeed, Dan Doel's proof is basically the generally proof just specialized to your particular functors and limits.

Berci's result is a special case of the fact that the external hom, $\mathcal C(X,-)$, is continuous. In fact, depending on how you define limits, this might be true by definition.

Answer 2

By definition of the terminal object, $\hom(X,T)$ is a one element set for every $X$.
And one element sets are exactly the terminal objects in $Set$.

Answer 3

Is $\hom(X,T)$ supposed to be the internal hom?

I'm going to change notation a bit, and write $[X,Y]$ for internal hom, $1$ for the terminal object, and $C(X, Y)$ for the external hom of category $C$.

I think this is pretty easy to show using representability definitions of things. So, a representability definition of an internal hom is:

$$C(-, [X, Y]) \cong C(- × X, Y)$$

The isomorphism is in the presheaf category $\hat{C}$. Now, if we consider $Y = 1$, then:

$$C(-, [X,1]) \cong C(- × X, 1) \cong C(-, 1)$$

The second isomorphism is the property Berci mentioned, all hom sets into the terminal object are singletons, so $C(Y × X, 1) \cong C(Y, 1)$. But this shows that $1$ and $[X,1]$ represent the same presheaf; they are isomorphic.

Answer 4

Yes, the internal hom object $[Y,T]$ is always the terminal object $T$ .

1. If $\mathcal{C}$ is a category with a terminal object $T$ and internal hom denoted by $[-,-]$, then by definition there is an isomorphism

   $$\mathcal C(X, [Y,T]) \cong \mathcal C(X\times Y,\; T)$$

   which is natural in $X$ and $Y$.

2. Because $T$ is terminal, the right-hand-side hom set $\mathcal C(X\times Y,\; T)$ has exactly one element; it's a singleton set. Hence there is an isomorphism of sets, for each $X$ and $Y$,

   $$\mathcal C(X, [Y,T]) \cong 1$$

3. In other words, the set $\mathcal C(X, [Y,T])$ is always a singleton set. Fixing $Y$ and allowing $X$ to vary, this shows that for any $Y$, the internal hom object $[Y,T]$ is the terminal object in $\mathcal{C}$ (i.e. it satisfies the universal property of the terminal object, having exactly one morphism from each object $X$), Q.E.D.

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[*] Here, $\mathcal{C}(-,-)$ denotes the external hom.