Intuition behind 'adjointness' of adjoint functor for a lay person

Question

For quite some time I, as a beginner in Category Theory, have been pondering about the intuition behind defining adjoint functors. I've gone through the other answers addressing this, but they seemed to be too technical for me. Is it right to think of adjoint functors to kind of 'translate a difficult problem to a simpler setting'? For example, in case of matrices, one can jump from one linear transform to another on the other side inside an inner product, if they are adjoint(although I haven't seen any instance of using adjointness to actually simplify things, I like to believe it, and I might be wrong).

Also, Ravi Vakil in his notes motivates adjoint functors in the following way:

Just as a universal property “essentially” (up to unique isomorphism) determines an object in a category (assuming such an object exists), “adjoints” essentially determine a functor (again, assuming it exists).

I'm not sure what he means by adjoints "essentially determine a functor".

How should I look at adjoint functors, and adjointness in general?

Answer 1

There are a lot of ways to understand adjoints, because they show up in many places, each of which gives a different perspective on things. For instance, if you're familiar with some galois theory or covering space theory, we have a close relationship between the lattice of field extensions and the lattice of subgroups of the galois group (resp. between the lattice of covering spaces and the lattice of subgroups of the fundamental group).

These facts are both extremely useful, since one half of the relationship (field extensions or covering spaces) is full of extremely complicated objects, and the other half of the relationship is "just" subgroups of a group (which is a discrete problem we can often solve completely!).

The "close relationship" between these lattices is mediated by a galois connection, which is actually just a historical name for an adjunction! Many properties of galois connections (which specialize to give the useful facts for galois theory, covering space theory, and many more) are actually properties of all adjunctions, or at least adjunctions between posets. These have their own name because we started studying galois connections in the 40s, but didn't even have a definition of adjunction until the mid 50s!

So one takeaway from this discussion is that yes. Adjunctions can help us study something complicated by transposing the complicated question to another domain where the question is easier to answer. I've gestured at some historical examples in the area of galois connections, but these ideas can be seen everywhere.

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Now, what about this "adjunctions specify a functor" business?

The idea here is that the left adjoint of a functor is unique. Here's a proof (crucially using yoneda's lemma):

Say that $L_1$ and $L_2$ are both left adjoints to $R$. Then for every $X$ and $Y$ we have

$$\text{Hom}(L_1 X, Y) \cong \text{Hom}(X, RY) \cong \text{Hom}(L_2 X, Y)$$

using the definition of an adjunction $L_1 \dashv R$ in the first isomorhpism, and the definition of an adjunction $L_2 \dashv R$ in the second isomorphism. So homming out of $L_1X$ and $L_2 X$ always give the same answer, which means $L_1 X \cong L_2 X$ by yoneda. Thus $L_1$ and $L_2$ are naturally isomorphic.

So this proof tells us that, if $R$ has a left adjoint (and of course the story is the same if it has a right adjoint by duality) that adjoint is (essentially) unique! Now there are LOTS of functors that are

1. easy to understand
2. have left adjoints that are hard to understand

For instance, the forgetful functor $U : \mathsf{Groups} \to \mathsf{Sets}$. This functor is super easy to understand -- it just takes a group $(G, \times, 1)$ and forgets everything but the underlying set $G$. One can show that this functor has a left adjoint (say, using one of the adjoint functor theorems), which sends a set $X$ to the free group on $X$!

The uniqueness of adjoints tells us that, in some sense, everything about the free group functor $F$ is already available inside the forgetful functor $U$. Moreover, just as in the case of galois connections, we can answer lots of interesting questions about the free group functor by transposing them along this adjunction to questions about the forgetful functor (which is much easier to study).

And, of course, there's nothing special about the free group functor here. There are countless examples of extremely complicated functors which are uniquely specified by being adjoint to some particularly simple functor. Again, this lets us answer questions about the complicated functor by transposing those questions to questions about the simple functor.

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I hope this helps ^_^

Answer 2

Here is a plausible explanation of the nomenclature, at least. I appreciate that this might not quite make sense without certain background, but maybe the shape of the explanation will still make sense.

One reason to regard adjoint functors as a type of "adjoint" is that in certain circumstances they are literally a categorification of the notion of adjoints one has in linear algebra. One categorical generalization of linear algebra is the notion of a triangulated category. I won't bother with the axioms here, but you should think of this as a structure somewhat similar to an Abelian category (though triangulated categories are almost never Abelian), where we have a notion of exact triangles $X\to Y\to Z$.

Associated to a triangulated category $\mathcal{C}$ is its Grothendieck group, $K_0(\mathcal{C})$, which is the free Abelian group on the objects of $\mathcal{C}$ quotiented by the relations $Y = X+Z$ for all exact triangles like above. The important thing here is that triangulated categories have their own notion of exact functor which takes exact triangles to exact triangles and preserves the shift functor $[1]$.

Crucially, an exact functor $F:\mathcal{C}\to \mathcal{D}$ of triangulated categories induces a homomorphism $F_*:K_0(\mathcal{C})\to K_0(\mathcal{D})$ of the Grothendieck groups.

Suppose given a $k$-linear triangulated category $\mathcal{C}$ which is Ext-finite in the sense that for all objects $E$ and $F$ of $\mathcal{C}$ one has $\{i\in \Bbb{Z}:\mathrm{Hom}_{\mathcal{C}}(E,F[i])\ne 0\}$ is a finite set. Then, one defines a bilinear form on $K_0(\mathcal{C})$ called the Euler form as follows: for all $X,Y\in \mathcal{C}$, $$ \chi_{\mathcal{C}}(X,Y):=\sum_{i\in \Bbb{Z}}(-1)^i \dim \mathrm{Hom}_{\mathcal{C}}(X,Y[i]). $$

Consider a pair of Ext-finite $k$-linear triangulated categories $\mathcal{C}$ and $\mathcal{D}$. If $F:\mathcal{C}\to \mathcal{D}$ and $G:\mathcal{D}\to \mathcal{C}$ are ($k$-linear) adjoint exact functors, $F\dashv G$, then for $X\in \mathcal{C}$ and $Y\in \mathcal{D}$, one has

$$ \chi_{\mathcal{D}}(FX,Y) = \sum_{i\in \Bbb{Z}}(-1)^i \dim \mathrm{Hom}_{\mathcal{D}}(FX,Y[i]) = \sum_{i\in \Bbb{Z}}(-1)^i\dim\mathrm{Hom}_{\mathcal{C}}(X,FY[i])= \chi_{\mathcal{C}}(X,GY). $$

So, in this case the induced maps $F_*:K_0(\mathcal{C})\to K_0(\mathcal{D})$ and $G_*:K_0(\mathcal{D})\to K_0(\mathcal{C})$ are adjoints in the classical sense with respect to the pairings $\chi_\mathcal{C}$ and $\chi_{\mathcal{D}}$.

An interesting example of this principle is in algebraic geometry when one considers a morphism of smooth and proper algebraic varieties $f:X\to Y$. There are induced (exact!) functors of their derived categories $\mathbf{R}f_*:\mathrm{D}^{\mathrm{b}}_{\mathrm{coh}}(X)\to \mathrm{D}^{\mathrm{b}}_{\mathrm{coh}}(Y)$ and $\mathbf{L}f^*: \mathrm{D}^{\mathrm{b}}_{\mathrm{coh}}(Y) \to \mathrm{D}^{\mathrm{b}}_{\mathrm{coh}}(X)$ such that $\mathbf{L}f^*\dashv \mathbf{R}f_*$. This one adjunction implies a lot of important results.

Answer 3

In addition to the other useful answers, for adjoint pairs of functors in suitable situations, left adjoints are right-exact, and right adjoints are left-exact. Thus, left adjoints have left-derived functors (some sort of homology), and right adjoints have right-derived functors (some sort of cohomology).

Also, there are some very-standard examples of adjoint pairs: for a group (not necessarily finite, but then with some continuity conditions) we have functors on $G$-modules $M$, the fixed submodule $M^G\to M$, and the cofixed quotient module $M\to M_G$. There is the adjunction $\mathrm{Hom}_G(M,N^G)\approx \mathrm{Hom}_G(M_G,N)$. Then the left-derived functors of the co-fixed vector functor are "group homology", and the right-derived functors of the fixed vector functor are "group cohomology". And instead of $G$-modules, we can consider $\mathfrak g$-modules for Lie algebras $\mathfrak g$. Or any other (nice) similar category.

The reason I mention that example is that (especially in older sources) group (and other) (co-) homology is "defined" in an ad-hoc fashion, using particular resolutions. Yeah, ok, those do turn out to be acyclic, and by the uniqueness... inescapably compute the derived functors.

Another big family is adjoints of forgetful functors. In the extreme case of the functor $L$ forgetting to "sets" (for concrete categories), the right adjoint $R$ forms the "free object". E.g., polynomial rings... In some slightly technical situations (Lie algebra repns to $\mathfrak g,K$-modules), the forgetfulness is only partial ... :) ...