# Pair of adjoint functors and showing that two certain maps are inverses

I'm currently self-studying "Categories and Sheaves"by Schapira and Kashiwara, and I've been stuck on problem 1.19 all day today, so I was hoping that someone could help me out.

Let $\mathcal{C}$, $\mathcal{C'}$ be categories and $L_v:\mathcal{C} \rightarrow \mathcal{C'}$, $R_v:\mathcal{C'} \rightarrow \mathcal{C}$ be functors such that $(L_v,R_v)$ is a pair of adjoint functors (v=1,2). Let $\epsilon_v: id_\mathcal{C} \rightarrow R_v L_v$ and $\eta_v: L_v \circ R_v \rightarrow id_{\mathcal{C'}}$ be the adjunction morphisms. Prove that the two maps $\lambda$,$\mu$ : $$Hom_{Fct(\mathcal{C},\mathcal{C'})}(L_1,L_2) \rightleftarrows^{\lambda}_{\mu} Hom_{Fct(\mathcal{C'},\mathcal{C})}(R_2,R_1)$$ given by : $$\lambda(\varphi):R_2 \rightarrow^{\epsilon_1 \circ R_2} R_1 \circ L_1 \circ R_2 \rightarrow^{R_1 \circ \varphi \circ R_2} R_1 \circ L_2 \circ R_2 \rightarrow^{R_1 \circ \eta_2} R_1$$ for $\varphi \in Hom_{Fct(\mathcal{C},\mathcal{C'})}(L_1,L_2)$ $$\mu(\psi) : L_1 \rightarrow^{L_1 \circ \epsilon_2} L_1 \circ R_2 \circ L_2 \rightarrow^{L_1 \circ \psi \circ L_2} L_1 \circ R_1 \circ L_2 \rightarrow^{\eta_1 \circ L_2} L_2$$ for $\psi \in Hom_{Fct(\mathcal{C}',\mathcal{C})}(R_2,R_1)$ are inverse to eachother.

So far, I've been trying to play around with the zig-zag identity, but it doesn't seem to lead me anywhere. Any help will be greatly appreciated!

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Update: Some progress today, but still nothing close to a solution. So far, I've been able to reduce it somewhat, but I'm not sure whether it is correct. – Shaf_math Dec 5 '11 at 17:43

Exercise $1.19$ of Categories and Sheaves, by Masaki Kashiwara and Pierre Schapira. Preview available at Google Books, and at Amazon.
For $i=1,2$ let $$L_i:\mathcal C\to \mathcal C',\quad R_i:\mathcal C'\to \mathcal C$$ be left and right adjoint functors, and let $$\varepsilon_i:\text{id}_{\mathcal C}\to R_iL_i,\quad\eta_i:L_iR_i\to\text{id}_{\mathcal C'}$$ be the adjunction morphisms. Let $X$ be in $\mathcal C$ and $X'$ be in $\mathcal C'$, and write $$a_i:\mathcal C'(L_iX,X')\to \mathcal C(X,R_iX')$$ for the adjunction isomorphism. Let $$\varphi:L_1\to L_2$$ be a morphism, and consider the diagram $$\begin{matrix} &a_2&\\ \mathcal C'(L_2X,X')&\to&\mathcal C(X,R_2X')\\ &&&\\ \varphi(X)^*\downarrow&&\downarrow b\\ &&&\\ \mathcal C'(L_1X,X')&\to&\mathcal C(X,R_1X'),\\ &a_1& \end{matrix}$$ where $b$ is defined by the requirement that the diagram commutes.
The goal is to find a morphism $$\psi:R_2\to R_1$$ such that $$b=\psi(X')_*.$$ Let $$f:X\to R_2X'$$ be a morphism. By the proof of Proposition [KS, $1.5.4$], we have $$a_2^{-1}f=\left[L_2X\xrightarrow{L_2f}L_2R_2X'\xrightarrow{\eta_2(X')}X'\right],$$ by which I mean that $a_2^{-1}f$ is equal to the composition between the brackets. This gives $$\varphi(X)^*a_2^{-1}f=\left[L_1X\xrightarrow{\varphi(X)}L_2X\xrightarrow{L_2f}L_2R_2X'\xrightarrow{\eta_2(X')}X'\right].$$ The functoriality of $\varphi$ yields $$\varphi(X)^*a_2^{-1}f=\left[L_1X\xrightarrow{L_1f}L_1R_2X'\xrightarrow{\varphi(R_2X')}L_2R_2X'\xrightarrow{\eta_2(X')}X'\right].$$ Using again the proof of Proposition [KS, $1.5.4$], we get $$a_1\,\varphi(X)^*a_2^{-1}f=$$ $$\left[X\xrightarrow{\varepsilon_1(X)}R_1L_1X\xrightarrow{R_1L_1f}R_1L_2R_2X'\xrightarrow{R_1\varphi(R_2X')}R_1L_2R_2X'\xrightarrow{R_1\eta_2(X')}R_1X'\right],$$ which, by functoriality of $\varepsilon_1$, is equal to $$\left[X\xrightarrow{f}R_2X'\xrightarrow{\varepsilon_1(R_2X')}R_1L_1R_2X'\xrightarrow{R_1\varphi(R_2X')}R_1L_2R_2X'\xrightarrow{R_1\eta_2(X')}R_1X'\right].$$ This implies the sought-for formula $$\psi(X')=\left[R_2X'\xrightarrow{\varepsilon_1(R_2X')}R_1L_1R_2X'\xrightarrow{R_1\varphi(R_2X')}R_1L_2R_2X'\xrightarrow{R_1\eta_2(X')}R_1X'\right].$$