Understanding adjoints has always been (and continues to be) a bit of a struggle for me.


Today I stumbled upon a property of adjoint functors which seemed extremely intuitive to me. I was wondering why this property isn't mentioned more often in introductory category theory literature, and whether or not it completely characterizes adjunctions.


If two functors $F:C\to D$ and $U:D\to C$ are adjoint $F\dashv U$, then for every $f:F(Y)\to X$ in $D$ there exists an $\hat f:Y\to U(X)$ in $C$ such that


$$ U(f)\circ \eta\_Y = \hat f$$
$$ \epsilon\_X\circ F(\hat f)=f$$


If we substitute the top equation into the bottom, we get


$$ \epsilon\_X\circ F(U(f)\circ \eta\_Y)=f$$


and by functoriality we get


$$ \epsilon\_X\circ F(U(f))\circ F(\eta\_Y)=f$$
$$ \epsilon\_X\circ (F\circ U)(f)\circ F(\eta\_Y)=f$$


What the last equation says is that we can recover any morphism $f$ from the action of the "round trip endofunctor" $F\circ U$ on it by pre-composing with $\epsilon\_X$ and post-composing with $F(\eta\_Y)$. These two morphisms are determined only by the domain and codomain of $f$ -- we only needed to know $X$ and $F(Y)$ in order to pick the two morphisms. We would have picked the same two morphisms for some $g\neq f$ as long as $g:F(Y)\to X$.


So, I believe it is correct to say that "if the domain of a morphism is within the range of a functor which has a right adjoint, then it can be recovered from the action of the composite endofunctor on it by pre-composition with some morphism and post-composition with some other morphism, where the choice of these two morphisms is completely determined by the domain and codomain of the original morphism". There is, of course, an equivalent statement for morphisms with a codomain in the range of a functor with a left adjoint.


So, my three questions are: (1) is this correct, (2) if so, why isn't it used to explain adjunctions to beginners (I certainly would have caught on quicker!) and (3) does the condition completely characterize adjoint functors?


Thanks,



"adjoint" =?= "inverse of composite endofunctor is uniform bi-composition"
(1) Yes. (2) Well, it doesn't give me any additional intuition. You didn't say why it helps you understand, so I can't judge what the advantage of it might be. I think this is really just a complicated way of giving the "bijection of hom-sets" condition.


(3) No, you need something more. For instance, let $r:B\to A$ be a surjection with section $s$, let $C$ have two objects $x$ and $y$ with $C(x,y)=B$, $C(y,x)=\emptyset$, and $C(x,x)=C(y,y)=1$ (only identities), let $D$ be similar using $A$ instead, and let $F:C\to D$ and $U:D\to C$ be the identity on objects and with action on arrows given by $r$ and $s$ respectively. Pick $\varepsilon$ and $\eta$ to be identities. Then every morphism in $D$ can be recovered, as you describe, but the components of $\eta$ are not natural, and the dual condition fails.


The "unknown (google)" comment above explained why if you additionally require the dual condition, plus naturality of $\eta$ and $\varepsilon$, then you do get an adjunction. (Although it's not clear to me from the condition you stated whether you wanted to require the morphism playing the role of $F(\eta)$ to actually be $F$ of something, which is also necessary for this argument to work.)


