Post History
#5: Post undeleted
by
clemens
·
2026-04-01T16:51:42Z (6 months ago)
#4: Post deleted
by
clemens
·
2026-04-01T16:50:41Z (6 months ago)
#3: Post edited
by
clemens
·
2026-04-01T16:47:07Z (6 months ago)
Let $f,g$ be any morphisms in $\text{Hom}(A,X)$. We want to show that, if $X^X$, $f=g$. 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 $f = π_1 \circ (f × g) = π_2 \circ (f×g) = g$. $\blacksquare$
- 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 $f = π_1 \circ (f × g) = π_2 \circ (f×g) = g$. $\blacksquare$
#2: Post edited
by
clemens
·
2026-04-01T16:46:31Z (6 months ago)
Let $f,g$ be any morphisms in $\text{Hom}(A,X)$. We want to show that, if $X^X$, $f=g$. 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 $f = π_1 \circ (f × g) = π_2 \circ (f×g) = g$. $\blacksquare$(I'm not experienced at category theory, so this is perhaps an overcomplicated proof.)
- Let $f,g$ be any morphisms in $\text{Hom}(A,X)$. We want to show that, if $X^X$, $f=g$. 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 $f = π_1 \circ (f × g) = π_2 \circ (f×g) = g$. $\blacksquare$
#1: Initial revision
by
clemens
·
2026-04-01T16:46:20Z (6 months ago)
Let $f,g$ be any morphisms in $\text{Hom}(A,X)$. We want to show that, if $X^X$, $f=g$. 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 $f = π_1 \circ (f × g) = π_2 \circ (f×g) = g$. $\blacksquare$
(I'm not experienced at category theory, so this is perhaps an overcomplicated proof.)