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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?

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If you are asking how to display the diagram on MSE then you could create a matrix such as $$\begin{matrix}A&\stackrel{f}{\rightarrow}&A\\\downarrow{g}&&\downarrow{h}\\C& \stackrel{i}{\rightarrow}&D\end{matrix}$$ using the following line $$\begin{matrix}A&\stackrel{f}{\rightarrow}&A\\\downarrow{g}&&\downarrow{h}\\C& \stackrel{i}{\rightarrow}&D\end{matrix}$$ – Giuseppe Jun 9 '12 at 11:43
I think what all of the above renditions of a commutative diagram indicate is that mathjax badly needs a mechanism for properly rendering commutative diagrams! – ItsNotObvious Jun 9 '12 at 14:13
@GiuseppeTortorella Cool, thank you very much! – Rudy the Reindeer Jun 10 '12 at 21:08
@GiuseppeTortorella That doesn't actually look so good. Can I make it any better somehow? – Rudy the Reindeer Jun 10 '12 at 21:15
If you know your way around xypic, then you can use Here's your diagram. – Bruno Stonek Jun 10 '12 at 21:34
up vote 3 down vote accepted

Given functors $F:\mathcal{D} \to \mathcal{C}$ and $G:\mathcal{D} \to \mathcal{C}$ with natural bijections $\text{hom}_\mathcal{C}(F(X),Y) \to \text{hom}_\mathcal{D}(X,G(Y))$ we say that $F$ is left adjoint to $G$. Thus tensor product is left adjoint to the Hom functor. I guess this naturally makes sense because in the defining equations the functor $F$ is on the left and $G$ on the right.

In terms of the proof, the map you have written down is correct, but of course one should actually show that everything works; i.e. that your map $\eta_{ab}$ is a bijective and that is is natural (i.e. that the diagram commutes).

It is also fairly standard to write $\text{Hom}_R(f \otimes \text{id}_M,g)$ as $(f \otimes \text{id}_M)^*$

Edit: Please see Bruno's comment below. For a map $f:A \to A'$ and a fixed $B$ it is normal to write $(f \otimes \text{id}_M)^*$. Otherwise $\text{Hom}_R(f \otimes \text{id}_M,g)$ seems to be the correct thing to write. (Note that in your question you need to change the $B$ in the lower left hand corner of the commutative diagram to a $B$')

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Nice, thank you very much! – Rudy the Reindeer Jun 10 '12 at 21:07
I'd be careful about your last sentence... If you write that, on the RHS you've lost all information about $g$, and that's not cool. Usually the superscript $*$ is used for the contravariant Hom on one variable, i.e. $h^*=\hom_R(h,M)$ where $M$ is a fixed module. That's not the case here, because $\text{Hom}_R(f \otimes \text{id}_M,g)$ is the hom bifunctor applied to a couple of maps. – Bruno Stonek Jun 10 '12 at 21:37
@Burno: Dear Bruno, you are correct. I was thinking we were doing naturality only for a map $f:A \to A'$ when I wrote that – Juan S Jun 10 '12 at 23:29
In the first sentence, shouldn't it be $G \colon \cal C \to \cal D$? – Joshua Taylor Sep 8 '13 at 2:33

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 $\mathcal C^{\text{op}} \times \mathcal D \longrightarrow \mathsf{Set}$.

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Nice, thank you very much! – Rudy the Reindeer Jun 10 '12 at 21:06
I think both these answers are very helpful. I accepted the other one because you have 2 upvotes and the other one only had one. – Rudy the Reindeer Jun 10 '12 at 21:08
It's ok:) just happy to help! – Paul Slevin Jun 10 '12 at 21:10

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