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I'm currently reading about adjoint functor (page 22 of these notes: In particular, it says that two functors $F: C\rightarrow C'$ and $G: C' \rightarrow C$ are adjoint if there exists an isomorphism of bifunctors: $Hom_{C'}(F(\circ),\circ) \cong Hom_C(\circ,G(\circ))$

By plugging in $F(\circ)$, we have: $Hom_{C'}(F(\circ),F(\circ)) \cong Hom_C(\circ,G(F(\circ)))$ as two functors from $C \rightarrow Set$. It then says that we have morphisms $X \rightarrow GF(X)$, functorial in X. I don't understand why it's functorial. So for each $X \in C$, we have $id_{F(X)} \in Hom_{C'}(F(X),F(X))$, which gives us a morphism $\varphi_X: X \rightarrow GF(X).$ If we have $f: X\rightarrow X'$, why the diagram commutes, in particular, why $GF(f) \circ \varphi_X = \varphi_{X'}\circ f ?$ Can anyone help me please.

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That's part of the definition of isomorphism of functors. – Zhen Lin Oct 20 '13 at 16:24
up vote 2 down vote accepted

Let's name $\psi \colon \text{Hom}_{C'}(F(\bullet),F(\bullet))\cong\text{Hom}_C(\bullet,G\circ F(\bullet))$ functorial isomorphism. In you notation so $\varphi_X = \psi(1_{F(X)})$.

Now the following equation holds $$\varphi_{X'} \circ f = \hom(f,G \circ F(X'))(\varphi_{X'})=\hom(f,G \circ F(X'))\circ \psi(1_{F(X)})$$ and this is also equal to $$\psi \circ \hom(F(f),F(X'))(1_{F(X')})=\psi(F(f))$$ in similar way we have that $$G\circ F(f) \circ \varphi_X = \hom(X,G\circ F(f))\circ \psi(1_{F(X)})$$ which is also equal to $$\psi \circ \hom(F(X),F(f))(1_{F(X)})=\psi(F(f))$$

From this it follows the equation you were looking for.

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