# Inituition of adjoint functors as best solution, toughest problem pairs

I'm trying to understand the use of adjoint functors and came across the interpretation of them as the optimal solution to some problem. But am trying to wrap my head around how to connect them with non-categorical concepts of optimisation.

From my current understanding this means if I have an adjoint pair $$F \dashv G$$ then the process prescribed by the associated universal morphisms from $$Y$$ to $$G$$ is the most efficient solution to the problem posed by $$G$$.

My question is how to take some functor $$G$$ and work out the meaning of the associated problem. Following the example used on the wikipedia page where $$G$$ is the forgetful functor from the category ring to rng. The associated optimisation problem seems to be, what is the fewest extra elements and relations I need to add to a rng so that I can label one of the elements as an identity and have it obey identity relations. However I feel like this phrasing of the problem isn't obvious when you just say $$G$$ is the forgetful functor.

Is there some process or heuristic so that I can take a functor $$G$$ and describe it as a optimisation problem in a less categoric way? Furthermore what kind of functors $$G$$ correspond to problems such as minimise this function?

• This might help : math.stackexchange.com/questions/20364/… – Arnaud D. Oct 22 at 15:43
• Not quite, my question is more about how to convert from optimisation (say in a physics sense) to adjoint fuctors and vice-versa, rather than how adjoint functors are inverses of each other. There might be something subtle in those answers but the connection is not obvious to me – N A McMahon Oct 23 at 9:54
• Honestly, I was confused by the exact same page a few days ago. It seemed like a few of the details of the example were inconsistent. Specifically, it seemed like the actual result would be the largest possible ring containing $R$. So, the example might have some issues. – Kevin P. Barry Oct 23 at 16:09
• @NAMcMahon Do you think my answer answers your question ? is there something unclear ? – jeanmfischer Oct 26 at 10:40
• @jeanmfischer I still need to go through it carefully, I suspect the poset generalisation interpretation was the part I was missing. However the rest of the text is dense enough I will need to sit down and carefully go through it (coming from physics) – N A McMahon Oct 26 at 11:26

Suppose we have a functor $$G : \mathcal{D} \to \mathcal{C}$$, and fix $$X$$ an object of $$\mathcal{C}$$. Then define the category $$X/\mathcal{D}$$ to be the category where objects are morphisms $$a : X \to G(Y)$$ for $$Y \in \mathcal{D}$$, and morphisms between $$a : X \to G(Y)$$ and $$b : X \to G(Z)$$ are morphisms $$f : Y \to Z$$ in $$\mathcal{D}$$ such that $$G(f) \circ a = b$$. Then the best solution to the problem posed by $$G$$ for $$X$$, if it exists, is the initial object in $$X/\mathcal{D}$$. Let's suppose it exists and call it $$\eta_X : X \to G(F(X))$$. Then you get that for any other map $$f : X \to G(Y)$$, by the fact that $$\eta_X$$ is initial, there is a unique map $$\overline{f} : F(X) \to Y$$, such that $$G(\overline{f}) \circ \eta_X = f$$.
To turn back to the poset analogy now, suppose you have an order preverving map between posets $$g : D \to C$$. Define, for $$x \in C$$ the poset $$x/D$$ of elements of $$y \in D$$ such that $$x \le g(y)$$ with the order induced by the one on $$D$$. The best solution, if it exists, is the smallest element of this poset. Suppose it exists and call it $$f(x)$$. We have of course that $$x\le g(f(x))$$ by definition. Also for any $$y$$ such that $$x\le g(y)$$, we have that $$f(x) \le y$$ also by definition of $$f(x)$$. If you can find one such $$f(x)$$ for every $$x$$ in $$C$$, you have built a map $$f : C \to D$$, note that this map if it exists is order preserving. The pair $$(f,g)$$ is called a Galois connection, and we have that for any $$x\in C$$ and $$y \in D$$, $$f(x)\le y$$ if and only if $$x \le g(y)$$.