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Let $F:A \leftrightarrows B:G$, $\theta:GF \cong 1_{B}$ and $\alpha:FG \cong 1_{A}$ be an equivalence of categories. I'm stuck trying to prove that if $A$ has all exponentials $Z^{Y}$, then so is $B$.

My attempt: I tried to prove that for every $Y$ in $B$, the product functor satisfies $-\times Y \cong F○(-\times G(Y))○G$, where this last has right adjoint for $F\dashv G$, $G\dashv F$, and the adjunctions can be "composed". However, I failed in finding such a natural isomorphism.

Could you help me?

John Mars
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First, observe that $\mathbf B$ has exponentials in the image of $F$, because $$\mathbf B(FX \times FY, FZ) \cong \mathbf A(X \times Y, Z) \cong \mathbf A(X, Z^Y) \cong \mathbf B(FX, F(Z^Y))$$ (observing that $F$ must be fully faithful and preserve products because it is an equivalence). However, since $F$ is an equivalence, it is essentially surjective, and so every object in $\mathbf B$ is in the image of $F$ up to isomorphism (which is sufficient, because universal properties are determined only up to isomorphism).

varkor
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