Motivated by the answer to this [question](https://mathoverflow.net/questions/407407/extending-representations-of-lie-subalgebras-to-the-whole-lie-algebra), I will ask the following question: Let $\mathcal{A}$ and $\mathcal{B}$ be small semisimple abelian categories. Let $U:\mathcal{A} \to \mathcal{B}$ be a functor that preserves and reflects exact sequences.


Is this enough in general to give an adjoint functor such that the unit of the adjunction is injective? If not, then what are examples of assumptions we can add to make this true?


Edit: In response to the comments below, the definition of semisimple is the one taken from nLab, so yes I am assuming finite direct sums.



> 
> An abelian category is called semisimple if every object is a
> semisimple object, hence a direct sum of finitely many simple objects.
> 
> 
> 


Edit: I am happy to assume that my categories are linear, that is enriced over vector spaces.



Adjoints of exact functors between semisimple abelian categories
This was too big to fit as a comment. Here is a cute, completely trivial, but incredibly useful fact.


Suppose $(f^\*, f\_\*)$ is an adjoint pair of functors between additive categories (not necessary abelian or anything of that nature).


If $X$ is an object such that $f^\*X \neq 0$, then the unit map $\eta\colon X \to f\_\*f^\*X$ is not zero. This is because this map occurs in the defining relation for unit/counit of an adjunction: the composition


$$f^\*X \to f^\*f\_\*f^\*X \to f^\*X$$


is the identity.


There is a dual result using $f\_\*$ too.


The relevance of this is that if, additionally, $X$ is a "simple object" (I leave it to you to pin the meaning of this down), then $\eta$ will be automatically injective. If in your category "every object is finite length", then induction will give injectivity on everything.


There are all sorts of fun games you can play with this completely trivial fact and get non-trivial results (for eg., you can deduce auto-equivalences at the level of derived categories using the complex $id \to f\_\*f^\*$ from just observing what the functors are doing at the $K\_0$ level - I am assuming a lot of things about the category in question, abelian, finite length, etc - so don't take this too literally).


I digress though. Coming back to the original question. The interesting bit about the question, in my opinion, is the existence of the adjoint in the first place. As I hope is clear from the above, it won't take much more additional hypothesis to make the unit of adjunction to be injective (or even an isomorphism - assume $f^\*$ preserves "simples").


In this vein, the "obstruction" to the adjoint existing has a lot to do with if the category "has enough objects". That's very vague, so let me amplify with the example I alluded to in the comments.


Consider the category $\mathcal{C}$ of finite dimensional representations for a semisimple Lie algebra over $\mathbb{C}$ ($\mathfrak{sl}\_2$ will do just fine). Every representation is completely reducible, the endomorphism ring of simples is $\mathbb{C}$, etc. - everything is about as nice as you can get.


Now consider the forgetful functor from $\mathcal{C}$ to vector spaces. This is about as nice as can be: exact, reflects exact sequences, etc. I don't think this has a left adjoint. Morally, such an adjoint would be the free module for the universal enveloping algebra (this is not finite dimensional at all) over a vector space (this is the adjoint in the big category of all modules).


This isn't a proof that such an adjoint can't exist, but should indicate the difficulty in producing one if there aren't "enough objects".


