To understand adjoint functors I tried to look at an example. Can you tell me if the following is correct?
Before I give the example I'd like to recap the definition: Given two categories $C,D$ and two functors $F: C \to D$ and $G: D \to C$ we say that $F$ and $G$ are adjoint if we can give a natural transformation isomorphism $\eta$ such that for every pair of objects $A \in \text{Obj}(C)$, $B \in \text{Obj}(D)$ and morphisms $f: A \to A^\prime$ in $C$ and $g: B \to B^\prime$ in $D$ the following diagram commutes:
$$
\begin{matrix}
\operatorname{Hom}(FA, B) & \xrightarrow{\eta_{AB}} & \operatorname{Hom}(A, GB) \\
\left\downarrow{\scriptstyle{\operatorname{Hom}(F(f), g)}}\vphantom{\int}\right. & & \left\downarrow{\scriptstyle{\operatorname{Hom}(f, G(g))}}\vphantom{\int}\right.\\
\operatorname{Hom}(FA^\prime, B^\prime)& \xrightarrow{\eta_{A^\prime B^\prime}} & \operatorname{Hom}(A^\prime, GB^\prime)
\end{matrix}
$$
I'm not sure whether $F$ is left adjoint to $G$ or the other way around. Which one is the left adjoint here?
And: is there a better way to display this diagram?
Now the example: We claim that $F = – \otimes_R M$ is the (left?) adjoint of $G = \operatorname{Hom}_R(M, -)$ where $M$ is an $R$-module. To see this we give a natural isomorphism $\eta_{A,B}$ (where $A,B$ are $R$-modules and $C = D = R-\textbf{Mod}$) such that the following diagram commutes:
$$\begin{matrix}\textrm{Hom}(A \otimes M, B)&\xrightarrow{\eta_{AB}}&\operatorname{ Hom}(A, \operatorname{Hom}(M,B))\\
\left\downarrow{\scriptstyle{\textrm{Hom}(f \otimes id_M, g)}}\vphantom{\int}\right.&&\left\downarrow{\scriptstyle{\textrm{Hom}(f, G(g))}}\vphantom{\int}\right.\\
Hom(A' \otimes M, B')&\xrightarrow{\scriptstyle{\eta_{A'B'}}}&\textrm{ Hom}(A^\prime, \operatorname{Hom}(M,B'))\end{matrix}$$
We define $\eta_{AB}$ to be the map $$\eta_{AB}: (f: a \otimes m \mapsto b) \mapsto (g: a \mapsto f(a \otimes -))$$
Then the diagram above commutes. Is this correct?
And is the downarrow map really $\operatorname{Hom}(f \otimes id_M, g)$? I didn't know what else to put there. And did I get the left/right adjointness the correct way around?
Best Answer
I will tell you how I remember if something is a left or right adjoint. Hopefully it's useful for you.
Let $\mathcal{C},\mathcal{D}$ be categories, and let $F:\mathcal C \to \mathcal D$, $G:\mathcal D \to \mathcal{C}$ be functors.
By definition $F$ is left-adjoint to $G$ if there are natural isomorphisms $$\overline{(\ )}:\mathcal{D}(FA, -) \to \mathcal{C}(A,G-)$$ $$ \overline{(\ )}:\mathcal{C}^{\mathrm op}(GB,-) \to \mathcal{D}^{\mathrm op}(B,F-) $$
for all objects $A \in \text{ob}\mathcal C$ and $B \in \text{ob}\mathcal D$, such that they are mutual inverses when you plug $B$ in the top one and $A$ in the bottom.
The way to remember that $F$ is a left adjoint is that in the first nice covariant natural transformation, $F$ is on the left.
So your diagram is simply the naturality square for the first transformation: hence $F$ is the left adjoint in that case.
EDIT OVER A YEAR LATER: An easier way to say the above is $F$ is left-adjoint to $G$ if there is a natural isomorphism $$ \mathcal D( F-_1, -_2) \cong \mathcal C(-_1, G-_2) $$ of functors $\mathcal C^{\text{op}} \times \mathcal D \longrightarrow \mathsf{Set}$.