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Adjunctions via Universal Arrows: Understanding a Proof.

I got stuck here while reading this question.

$g=(g^♭)^♯$

How can I derive this equation?

confusingmaster
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1 Answers1

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For any morphism $f : A \to UY$, the adjunct $f^\sharp$ is defined to be the unique morphism $f^\sharp : FA \to Y$ such that $U(f^\sharp) \circ \eta_A = f$.

If $g: FA \to Y$ then $g^\flat: A \to UY$ is defined to be $Ug \circ \eta_A$. Then $(g^\flat)^\sharp$ is the unique morphism such that $U((g^\flat)^\sharp) \circ \eta_A = g^\flat$. Since $Ug\circ \eta_A = g^\flat$, by uniqueness we have $g = (g^\flat)^\sharp$.

Matthew Towers
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  • Thank you!! I didn't know it was so difficult to prove the uniqueness of morphism. As a beginner in category theory, is there any books or papers that I can refer to? – confusingmaster Sep 17 '22 at 02:43
  • there's no proof of the uniqueness of any morphism here - you should read it again. The uniqueness follows from the hypothesis of the theorem in the question you linked (it comes from the notion of a universal arrow which is defined in section 3 of the text linked in that question). – Matthew Towers Sep 17 '22 at 12:02