In category theory why is a right adjoint not a left adjoint?
You ask:
If you do have [a bijection between morphisms FA → B and morphisms A → GB], then could you not also take your bijection 'in the other direction' between morphisms GA → B in C and morphisms A → FB in D [...]?
This argument doesn’t quite work, since taking the bijection in the other direction, it will go between morphisms A → GB and morphisms FA → B. The functor G still only appears on the codomains of morphisms.
To give a concrete example where no such bijection exists: let (F, U) be the “free group”/“underlying set” functors between Set and Gp.
Now, F is left adjoint to U. But they can’t be adjoint the other way round! If they were, that would give a bijection between $\mathbf{Sets}(U1,\phi)$ and $\mathbf{Gp}(1,F\phi)$ (where $1$ denotes the trivial group, and $\phi$ the empty set). But $\mathbf{Sets}(U1,\phi)$ is empty, while $\mathbf{Gp}(1,F\phi)$ has one element, since $F\phi \cong 1$.
An intuition for adjoints? There’s no easy, one-size-fits-all answer; but a good place to start is with these sorts of free/forgetful examples. Typically one can thing of a left adjoint as adding stuff, as freely as possible — perhaps new elements, perhaps some structure, perhaps imposing some equations if necessary. On the other hand, a right adjoint typically forgets things — forgets structure, sometimes perhaps throws away elements too...
The point you mention that they're a generalisation of inverses is also a good one. The Stanford Encyclopedia of Philosophy calls them “conceptual inverses”, which depending on how you feel about philosophy may be very helpful or not at all.
But really the best way to get intuition for adjoints comes from looking at as many examples as possible; not necessarily all in a hurry, but every now and then, for a while. They’re one of those concepts that doesn't usually come quickly (it didn't for me, nor for anyone I've seen learning category theory), but which — if you give it time to percolate, and occasional exercises — will sooner or later “click” and suddenly seem so natural you can't imagine not understanding it.
Incidentally, a similar question was also asked some time ago at mathoverflow. I very much like the current second answer, giving a rather different example of adjoints: viewing the posets Z and R as categories in the usual way (i.e. there's a unique map x → y whenever x ≤ y), the “ceiling” function R → Z is left adjoint to the inclusion Z → R, while dually the “floor” function is right adjoint to the inclusion. This suggests the slogan: left adjoints round up (again, adding just as much as is needed to get an integer); right adjoints round down (forgetting the non-integer part).
If you're saying that a bijection between sets of morphisms
$$ \mathcal{D} (FA, B) = \mathcal{C}(A, GB) $$
implies necessarily a bijection
$$ \mathcal{D} (B, FA) = \mathcal{C}(GB, A) $$
for every pair of functors $F$ and $G$, then you should prove it. Shouldn't you? -"Why not?" is not a proof.
Anyway, I'll try to show you "why not".
Because there are examples of particular functors $F$ and $G$ where you have the first bijection without having the second one.
Take for instance the first example of adjoint functors in Mac Lane's book:
$$ V: \mathbf{Set} \leftrightarrows \mathbf{Vct_k} : U \ . $$
Here
- $\mathbf{Set}$ is the category of sets.
- $\mathbf{Vct_k}$ is the category of $\mathbf{k}$-vector spaces, $\mathbf{k}$ a field.
- $U$ is the forgetful functor which sends every vector space $W$ to its underlying set of vectors $U(W)$.
- $V$ is the functor which sends every set $X$ to the vector space $V(X)$ with basis $X$.
This pair of functors are adjoint: $U$ right, $V$ left. You have an easy pair of bijections, inverse one to each other:
$$ \varphi : \mathbf{Vct_k}(V(X), W) \rightleftarrows \mathbf{Set}(X, U(W)): \psi $$
defined as follows:
- For each function $g: X \longrightarrow U(W)$, $\psi (g) : V(X) \longrightarrow W$ is the unique linear map which extends the function $g$.
- For each linear map $f: V(X) \longrightarrow W$, $\varphi (f) : X \longrightarrow U(W)$ is the restriction $f_{\mid X}$ of $f$ to the set $X$.
You can check that both compositions $\varphi \circ \psi$ and $\psi \circ \varphi$ are identities.
Now, if your guess was right, you should be able to produce another pair of bijections
$$ \varphi' : \mathbf{Vct_k}(W, V(X)) \rightleftarrows \mathbf{Set}(U(W), X): \psi' $$
Can you?
EDIT. Maybe I should have warned you NOT to try it. Here is a little example that shows that this is impossible (and hence $U$ is a right adjoint to $V$ and $V$ a left adjoint to $U$, but NOT the other way around).
Take $\mathbf{k} = \mathbb{R}$, the real numbers, $W = \mathbb{R}$ and $X = \left\{ *\right\}$ a set with a single element. Then,
- $V(*) = \mathbb{R}$.
- $\mathbf{Vct}_{\mathbb{R}} (W, V(X)) = \mathbf{Vct}_{\mathbb{R}}(\mathbb{R}, \mathbb{R}) = \mathbb{R}$.
- $\mathbf{Set}(U(W), X) = \mathbf{Set}(\mathbb{R}, *) = \left\{ *\right\}$.
So, there is no possible bijection the other way around. Hence, $U$ cannot be left adjoint to $V$ and $V$ cannot be right adjoint to $U$.