In How to glue perverse sheaves Beilinson claims that the category of perverse sheaves on the complex unit disk $D$ with the stratification with the closed strata $\{0\}\subset D$ is equivalent to the category $C$ of diagrams $X \underset{u}{\stackrel{v}{\rightleftarrows}} Y$ with $X,Y$ some $\mathbb{C}$-vector spaces such that $\text{id}_X-uv$ and $\text{id}_Y-vu$ are both invertible.
Let $(V,f)^0$ for $f\in \text{End}(V)$ be the maximal subspace of $V$ on which $f$ acts as a nilpotent. I (pretend to) understand why the category of perverse sheaves is equivalent to the category $C'$ with the objects given by tuples $(X,Y,\phi,u,v)$ with $\phi\in\text{Aut}(Y)$, $v:X\to Y,$ $u:Y\to X$, and such that there is a diagram
$$X{\stackrel{v}{\rightarrow}}(Y,\text{id}_Y-\phi)^0{\stackrel{u}{\rightarrow}}X$$
with $v=\text{id}_Y-\phi.$ In particular, $v$ must actually land in $(Y,\text{id}_Y-\phi)^0$.
Beilinson gives a functor from $C'\to C$, which takes a diagram $X \underset{u}{\stackrel{v}{\rightleftarrows}} Y$ to the tuple $((X,uv)^0,Y,\text{id}_Y-vu,u,v).$ Could somebody help me figure out why this functor is an equivalence?
I suspect that the inverse equivalence should look something like this: given the spaces $X,Y$, we define $Y'$ as the quotient of $Y$ by $(Y,\text{id}_Y-\phi)^0$ and take a diagram $X\oplus Y' {\rightleftarrows} Y.$ But I couldn't figure out what the maps should be. Or maybe I am just completely wrong and looking in the wrong place?
EDIT: I think I figured it out, I am going to type it up in a couple hours to get the question out of the unanswered queue.
Best Answer
Assume we have the data $(X,Y,\phi,u,v)$. Notice that $Y$ splits as a direct sum $(Y,\text{id}_Y-\phi)^0\oplus \text{im}(\text{id}_Y-\phi)^n$ for $n$ sufficiently large and denote this stable image $Y'$. Note that $\text{id}-\phi$ restricts to an automorphism $\psi$ on $Y'$. Let $\psi$ be this restriction.
Then we define $\tilde{X}=X\oplus Y'$, $\tilde{Y}=Y=(Y,\text{id}_Y-\phi)^0\oplus Y'$ and define $\tilde{v}=v \oplus \psi:X\oplus Y'\to (Y,\text{id}_Y-\phi)^0\oplus Y'$ and $\tilde{u}=u\oplus\text{id}_{Y'}:(Y,\text{id}_Y-\phi)^0\oplus Y'\to X\oplus Y'$.
Then $(\tilde{X},\tilde{Y},\tilde{u},\tilde{v})$ is an object of category $C$, since $\text{id}_\tilde{X}-\tilde{u}\tilde{v}=(\text{id}_\tilde{X}-uv)\oplus\phi|_{Y'}$ is something invertible (and this is exactly what I missed early on), and $\text{id}_\tilde{Y}-\tilde{v}\tilde{u}=\phi\in\text{Aut}(\tilde{Y})$. This construction is functorial and is the inverse to the functor given by Beilinson.