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The exercise is the following: Suppose $F$ has right adjoint $G$ and $H$ has right adjoint $J$ ($F:\mathbb{A}\rightarrow\mathbb{B}$ and $H:\mathbb{B}\rightarrow\mathbb{C}$ with $\mathbb{A,B,C}$ categories). Prove that $GJ$ is the right adjoint of $HF$.

What is the best way to start with? I want to do this by the two-way-rule, thus by bijective correspondence, but i don't see how to start. Someone who can help me?

Thank you very much.

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$$\hom_{\Bbb A}(a,\ GJ(c))\, \simeq\, \hom_{\Bbb B}(F(a),\ J(c))\, \simeq\, \hom_{\Bbb C}(HF(a),\ c)\,.$$

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  • $\begingroup$ Oh yes, that is also a definition of adjoints. Thanks. Do you have also ideas about proving by two-way rule? $\endgroup$
    – HamOpCat
    Commented Mar 2, 2014 at 12:35
  • $\begingroup$ What is the two-way rule? $\endgroup$
    – Martin Brandenburg
    Commented Mar 2, 2014 at 12:40
  • $\begingroup$ Given an arrow $f:HF(X)\rightarrow Y$ find $\bar{f}:X\rightarrow GF(Y)$ and also for $g$ find a $\bar{g}$. Both new functions must have the property that $\bar{\bar{f}}=f$ and $\bar{\bar{g}}=g$ $\endgroup$
    – HamOpCat
    Commented Mar 2, 2014 at 13:20

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