The simplest example of Koszul duality (see introduction of [Beilinson, Ginzburg, and Soergel - Koszul Duality Patterns in Representation Theory](https://doi.org/10.1090/S0894-0347-96-00192-0)) is as follows.


Let $V = \mathbb{C}x$ be a $1$ dimensional vector space. Then the exterior algebra is $A=\mathbb{C}[x]/(x^2)$, and the symmetric algebra is $A^! = \mathbb{C}[x]$. Koszul duality states that there exists an isomorphism (where "mod" means the category of finitely generated modules):


$F: D^b(A\mbox{-mod}) \rightarrow D^b(A^!\mbox{-mod})$


The heart of the category on the left is $A\mbox{-mod}$; what is the image of this on the right? It should be possible to work this out by working through the above paper, but there are many things in that paper which I don't understand (and this toy example should be easy).



Koszul (exterior/symmetric) duality for a 1-dim vector space
"mod" has to mean the category of finitely generated *graded* modules, if not, there is no such equivalence.


Up to graded shifts (and isomorphism), the category $A\text-\mathsf{mod}$ in your example only has two indecomposable objects: the trivial module $\mathbb C$ and the length two projective-injective module $A$. Let $I$ denote the indecomposable graded injective module cogenerated in degree $0$, so $I=A \langle -1 \rangle$.


Theorem 2.12.5 in [Beilinson, Ginzburg, and Soergel - Koszul Duality Patterns in Representation Theory](https://doi.org/10.1090/S0894-0347-96-00192-0) contains all you need in order to understand the functor $F$ in this example. According to Theorem 2.12.5, $$F(\mathbb C \langle-n\rangle )=A^! [n] \langle n \rangle$$ and $$F(I \langle-n\rangle)=\mathbb C [n] \langle n \rangle .$$
In conclusion, the indecomposable objects in the essential image $F(A\text-\mathsf{mod})$ are $\{ A^! [n] \langle n \rangle \}\_{n \in \mathbb Z}$ and $\{\mathbb C [n] \langle n \rangle \}\_{n \in \mathbb Z}$.


A typical exact sequence $$0 \rightarrow \mathbb C \rightarrow I \rightarrow \mathbb C\langle-1\rangle \rightarrow 0$$ in $A\text-\mathsf{mod}$ is sent by $F$ to the exact sequence $$0 \rightarrow A^! \rightarrow \mathbb C \rightarrow A^! [1] \langle 1 \rangle \rightarrow 0$$ in $F(A\text-\mathsf{mod})$.


