Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

$$ \begin{matrix} \operatorname{Hom}(FA, B) & \xrightarrow{\eta_{AB}} & \operatorname{Hom}(A, GB) \\ \left\downarrow{\scriptstyle{\operatorname{Hom}(F(f), g)}}\vphantom{\int}\right. & & \left\downarrow{\scriptstyle{\operatorname{Hom}(f, G(g))}}\vphantom{\int}\right.\\ \operatorname{Hom}(FA^\prime, B)& \xrightarrow{\eta_{A^\prime B^\prime}} & \operatorname{Hom}(A^\prime, GB^\prime) \end{matrix} $$


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?

share|improve this question
2  
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 Tortorella 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! –  Matt N. Jun 10 '12 at 21:08
    
@GiuseppeTortorella That doesn't actually look so good. Can I make it any better somehow? –  Matt N. 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
add comment

2 Answers 2

up vote 2 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 $B$')

share|improve this answer
    
Nice, thank you very much! –  Matt N. 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 *bi*functor 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
add comment

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

share|improve this answer
    
Nice, thank you very much! –  Matt N. 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. –  Matt N. Jun 10 '12 at 21:08
1  
It's ok:) just happy to help! –  Paul Slevin Jun 10 '12 at 21:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.