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I've noticed that in many cases, whenever a functor $G:\mathscr B\to\mathscr A$ has a left adjoint, there's some kind of universal property around (I don't know how to state this more precisely other than using the word "around"). Examples (without explanation):

  • $G: \mathbf{Vect}_k\to \mathbf{Set}$ is the forgetful functor;
  • $G: \mathbf{Grp}\to \mathbf{Set}$ is the forgetful functor;
  • $G: \mathbf{Ab}\to \mathbf{Grp}$ is the functor that forgets that an abelian group is abelian;
  • $G: \mathbf{Set}\to \mathbf{Top}$ is the functor that assigns to a set the corresponding topological space with the indiscrete topology;
  • $G: \mathbf{Top}\to \mathbf{Set}$ is the forgetful functor.

So I was wondering whether there's some general connection between universal properties and the existence of left adjoints. This question probably is too imprecise, but here's what I got when trying to generalize the above examples:

Suppose $F: \mathscr A \to \mathscr B$, $G:\mathscr B\to \mathscr A$, and $F$ is left adjoint to $B$. Then for all $A\in\mathscr A, B\in\mathscr B$, $$\mathscr B(F(A),B)\simeq \mathscr A(A,G(B)).$$

In this case, my conjecture is that there is the following universal property:

For any $\phi\in \mathscr A(A,G(B))$, there exists a unique $\overline \phi\in \mathscr (F(A),B)$ such that the following diagram commutes:

enter image description here

When $G$ is one of the functors from the above example, it's always clear what $\pi$ is, but in the general case I don't really see what the arrow $\pi$ would be. Is there a way to define it in general so that the diagram commutes? If my conjecture is true, I would call the quoted part "the general (case of) universal property" (not sure if this is standard or not). And it would follow that whenever a functor has a left adjoint, there is an associated universal property.

Apparently, the converse of my conjecture is true (if there's a way to define $\pi$ that I'm confused about) because if there's such universal property, one could define the isomorphism from the definition of adjunction by $\phi\mapsto \overline\phi$ and in the other direction by $\psi\mapsto G(\psi)\circ \pi$. (Let me know if this is not true.)

Is there more stuff that can be said about the connection of left adjoints and universal properties?

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If you look at the Wikipedia article article on adjoint functors, one of the three equivalent definitions of an adjoint pair uses universal properties. To answer one of your specific questions: your map $\pi : A \to GF(A)$ is the unit of the adjunction evaluated at $A$ and is given explicitly by applying the hom-set adjunction to the identity map $1_{F(A)} : F(A) \to F(A)$.

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  • $\begingroup$ Interesting, I'm reading Leinster's book, and I don't think he draws the connection between adjoint functors and universal properties. The two equivalent definitions of adjunctions he gives is via units/counts and via initial objects in the comma category. I don't find those two definitions intuitive at all, whereas the definition that uses universal properties looks very natural (and I even partially came up with it myself). $\endgroup$ – user634426 Mar 20 at 3:30
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    $\begingroup$ @user634426 The universal property you stated in your question means precisely that the unit is initial in the appropriate comma category. If you look at section 2.3, he uses the universal property (of your first example) to introduce the topic, and he states at the end (last paragraph before the exercises that adjunctions are equivalent to the universal property. $\endgroup$ – Arnaud D. Mar 20 at 9:21
  • $\begingroup$ @ArnaudD. Thanks for pointing that out. I didn't pay enough attention to the connections between initial objects in a suitable comma category and universal property (and I still think that this could have been done more explicitly). Do you know why Leinster doesn't prove the equivalence of hom-definition to the comma-definition directly, but instead uses the triangle identities as an intermediate step? I haven't tried to prove it directly (I'm going to), but it seems to me that the direct proof would be more intuitive. $\endgroup$ – user634426 Mar 21 at 4:05
  • $\begingroup$ @ArnaudD. To follow up on my previous comment, here is my attempt to prove the equivalence directly. $\endgroup$ – user634426 Mar 21 at 18:45

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