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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2002

Paper 20

Using the transpose formulas of the preceding part,
$$
\varepsilon_{FC}F(\eta_C)=\alpha^{-1}_{C,FC}(\alpha_{C,FC}(1_{FC}))=1_{FC}.
$$
Similarly,
$$
G(\varepsilon_D)\eta_{GD}=\alpha_{GD,D}(\alpha^{-1}_{GD,D}(1_{GD}))=1_{GD}.
$$
Thus the <adjunction unit> and <adjunction counit> satisfy \b[both <triangle identities for an adjunction>],
$$
\boxed{\varepsilon_F\circ F\eta=1_F,\qquad G\varepsilon\circ\eta_G=1_G.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 20 / 6 / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 20 / 6 / ii / Solution

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Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 20 / 6 / ii / Solution

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