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My question is how can we see that if $F$ and $F'$ are both left adjoints of $G$, there is natural isomorphism between $F$ and $F'$? So how can we show that adjoints are unique upto isomorphism?

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up vote 7 down vote accepted

If $F \vdash G$, $F' \vdash G$, then we have units/counits $\epsilon:I \rightarrow FG, \eta:GF \rightarrow I$, and $\epsilon':I \rightarrow F'G, \eta':GF' \rightarrow I$. Then $\epsilon F':F' \rightarrow FGF'$, while $F\eta' :FGF' \rightarrow F$. Composing these gives a natural transformation $F \rightarrow F'$. Now you, of course, has to find the inverse map and show it actually is an inverse. The nice thing about this approach is it's already natural! For me, the key is to understand how to translate your information into the "algebra of functors" and work with it there.

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This approach is quite cumbersome. But it works in arbitrary $2$-categories, not just $\mathsf{Cat}$. – Martin Brandenburg Nov 24 '13 at 23:38

We have $\hom(Fx,y) \cong \hom(x,Gy) \cong \hom(F'x,y)$ natural in $x,y$, hence $F \cong F'$ (Yoneda).

PS: Compare this with the proof that adjoint operators between Hilbert spaces are unique.

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I had been wondering how to show that we get an isomorphism between the functors $F$ and $F'$, since applying the Yoneda Lemma at the individual $x$ gives only an iso $Fx\simeq F'x$, but we want a natural one. I figured out that the $1_{Fx}$ on Hom$(Fx,Fx)$ corresponds to $\epsilon'\circ F'\eta$, and this is natural in $x$ which is natural. – Stefan Hamcke Jan 25 '14 at 21:53
On the other hand, I like your approach in…, as it doesn't mention at all the unit and counit but only uses the naturality in both coordinates, and is much more abstract overall. – Stefan Hamcke Jan 25 '14 at 21:55

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