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I'm following Turi's Category Theory Lecture Notes, which can be found online. I'll try to make downloading a copy unnecessary. Here's my problem.

Definition: An adjunction $\langle F\dashv U, \varphi\rangle$ from a category $\mathbf{C}$ to a category $\mathbf{D}$ is given by a pair of opposite functors $F: \mathbf{C}\to\mathbf{D}\,$ & $\, U:\mathbf{D}\to\mathbf{C}$ and, for any $A$ is $\mathbf{C}$ and any $Y$ in $\mathbf{D}$, a natural bijection $$\varphi_{A, Y}:\mathbf{D}(FA, Y)\to\mathbf{C}(A, UY)$$ in the sense that, for every $g: FA\to Y$ two conditions hold:

(1) For every $k:Y\to Y'$, $\varphi (k\circ g)=Uk\circ\varphi (g)$ and

(2) For every $h:A'\to A$, $\varphi (g\circ Fh)=\varphi (g)\circ h$.

Lemma: The naturality conditions for an adjunction $\langle F\dashv U, \varphi\rangle$ are equivalent to the following: for every $f: A\to UY$

(1) For any $k: Y\to Y'$, $\varphi^{-1}(Uk\circ f)=k\circ\varphi^{-1}(f)$ and

(2) For any $h:A'\to A$, $\varphi^{-1}(f\circ h)=\varphi^{-1}(f)\circ Fh$.

My "proof": Put $g=\varphi^{-1}(f)$ in the definition so that $f=\varphi (g)$, which we can do by the fact that $\varphi_{A, Y}$ is a bijection. Then $$\varphi (k\circ g)=Uk\circ\varphi (g)\Leftrightarrow k\circ g=\varphi^{-1}(Uk\circ\varphi (g))\Leftrightarrow k\circ\varphi^{-1}(f)=\varphi^{-1}(Uk\circ f)$$ and similarly $\varphi (g\circ Fh)=\varphi (g)\circ h\Leftrightarrow\varphi^{-1}(f)\circ Fh=\varphi^{-1}(f\circ h)$. QED.

Is this correct? How would you do it?

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This is from p.38, ibid. The theorem just after rests on this lemma, but its proof uses similar reasoning and I think there's some danger of circular reasoning thanks to my proof. I'll explain myself more later if I have time. – Shaun Nov 11 '13 at 16:10
Instead of asking us if this is correct, you should ask this yourself. In particular the last $\Leftrightarrow$ needs some reasoning. – Martin Brandenburg Nov 11 '13 at 16:24
I have. I don't think it's correct but, when I try to come up with another proof, this is all I get. The very last $\Leftrightarrow$ (after the "similarly") has been written out explicitly and, yeah, it's similar; I'll type that out later too. I need to think it all through again. – Shaun Nov 11 '13 at 16:36
@Shaun you proof seems correct to me. – Giorgio Mossa Nov 13 '13 at 10:22
@Giorgio: Thank you. My problem is with a theorem later on then. I've followed this proof around to check that the subscripts make sense and so on. I just wanted to be sure. – Shaun Nov 13 '13 at 13:12
up vote 2 down vote accepted

Try to prove the following statement. Your question about adjunctions is a special case of this.

Let $F,G : C \to \mathsf{Set}$ be set-valued functors. Let $\alpha : F \to G$ be a family of bijective maps $\alpha(c) : F(c) \to G(c)$ for all $c \in C$. Then $\alpha$ is natural iff $\alpha^{-1}$ (defined by $\alpha^{-1}(c) = \alpha(c)^{-1}$) is natural.

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Thank you, @Martin. I'm not having much luck with this so far, although (I think) I see the connection. I'm not sure how to work with $\alpha (c)^{-1}$ though. I'll come back to this; it looks interesting :) – Shaun Nov 13 '13 at 14:31

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