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From Ravi Vakil, Fundations of Algebraic Geometry.

I want to ask if anyone can give a hint in how to prove Execrise 1.5.B(page 43). I tried to draw the diagram for half an hour but the resulting morphisms does not seem commute at all. Nor could I prove any natural bijection must arise from some unit/counit this way.

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up vote 2 down vote accepted

You might try $\eta_A := \tau_{A,FA} (1_{FA})$.

Here is the "trick": Take any $g \colon FA \to B$. Because $\tau$ is natural in the second variable, we have commutative diagram:

$$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \begin{array}{llllllllllll} Mor(FA,FA) & \ra{\tau_{A,FA}} & Mor(A,GFA) & \\ \da{g\,\circ\,\_} & & \da{Gg \, \circ\,\_} \\ Mor(FA,B) & \ra{\tau_{A,B}} & Mor(A,GB) \\ \end{array}$$

Now, take $1_{FA}$ from upper left corner, and evaluate both paths in it. You get $Gg \circ \eta_A = \tau_{A,B}(g)$, as desired.

For $\epsilon$, proof is dual.

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Thanks! I need to think about it... – Bombyx mori Feb 13 '13 at 16:44

This is the description of adjoints via unit and counit. It follows immediately from the Yoneda Lemma. Details can be found in every introduction to category theory, for example: Mac Lane, Categories for the Working Mathematician, Chapter IV, Section 1, Theorem 1.

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Yeah, but I think it is a fun exercise. – rafaelm Feb 13 '13 at 18:29
Sure, one has to do this on its own. I just wanted to add the reference. – Martin Brandenburg Feb 13 '13 at 21:59

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