Comments on Prove that if $X^X$ is a terminal object then $X \to \mathbf{1}$ is a monomorphism
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Prove that if $X^X$ is a terminal object then $X \to \mathbf{1}$ is a monomorphism
I want to know how to solve the following exercise in the textbook Conceptual Mathematics by Lawvere [Session 31, Exercise 2].
Let $X$ be an object in a cartesian closed category. Show that the following two properties are equivalent:
- $X \to \mathbf{1}$ is a monomorphism;
- $X^X = \mathbf{1}$.
I see that (1) is the same as saying that for all objects $A$, there is at most one map $A\to X$. Using this, I can easily prove that (1) implies (2), but the other direction eludes me.
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| User | Comment | Date |
|---|---|---|
| Hernán Ibarra Mejia | (no comment) | Apr 1, 2026 at 16:53 |
Let $f,g$ be any morphisms in $\text{Hom}(A,X)$. We want to show that, if $X^X=1$ then $f=g$. We do this by showing that the projection maps $π_1, π_2: X×X → X$ are equal: $π_1, π_2 \in \text{Hom}(X×X,X) = \text{Hom}(X,X^X) = \text{Hom}(X,1) \sim 1$. Hence $f = π_1 \circ (f × g) = π_2 \circ (f×g) = g$. $\blacksquare$
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