This is something I've never understood.

Let me first recall the classical case, in which you start with a smooth algebraic variety $X$ over $\mathbb{C}$. One has the algebraic de Rham complex $\Omega^\bullet_{X/\mathbb{C}}$ and its analytification $\Omega^\bullet_{X^{an}}$ on $X^{an}$. Then:

**Theorem (Grothendieck)**. The morphism of complexes $\Omega^\bullet_X \to \Omega^\bullet_{X^{an}}$ induces isomorphisms at the level of hypercohomology $$\mathbb{H}(X, \Omega^\bullet_{X/\mathbb{C}}) \to \mathbb{H}(X^{an}, \Omega^\bullet_{X^{an}}),$$ the first one being with respect to the Zariski topology while the second is taken in the usual cohomology of the complex analytic variety $X^{an}$.

Now assume you are given an integrable connection $$ \nabla: \mathcal{F} \to \mathcal{F} \otimes_{\mathcal{O}_X} \Omega^1_X $$ on some locally free $\mathcal{O}_X$-module. This defines a de Rham complex $DR(\mathcal{F}, \nabla)$ and one can also consider the analytic version $DR(\mathcal{F}^{an}, \nabla^{an})$. There is still a map $$ \mathbb{H}^\ast(X, DR(\mathcal{F}, \nabla)) \to \mathbb{H}^\ast(X^{an}, DR(\mathcal{F}^{an}, \nabla^{an}) $$ but now very easy examples show that this won't be an isomorphism in general. That's where the assumption of regular singularities comes in.

Fix a good compactification $j: X \hookrightarrow \bar{X}$ (so $\bar{X}$ is smooth and the complement is a normal crossings divisor $D$). The $(\mathcal{F}, \nabla)$ has regular singularities if it can be extended to some coherent $\mathcal{O}_{\bar{X}}$-module $\bar{\mathcal{F}}$ equipped with a logarithmic integrable connection $$ \overline{\nabla}: \bar{\mathcal{F}} \to \bar{\mathcal{F}} \otimes \Omega^1_X(\log D) $$ Now the advantage is that one has the GAGA theorem at disposal, so $$ \mathbb{H}^\ast(\bar{X}, DR(\bar{\mathcal{F}}, \bar{\nabla})) \simeq \mathbb{H}^\ast(\bar{X}^{an}, DR(\bar{\mathcal{F}}^{an}, \bar{\nabla}^{an})) $$ for any such extension. The problem is of course that the natural restriction map $$ \mathbb{H}^\ast(\bar{X}, DR(\bar{\mathcal{F}}, \bar{\nabla})) \to \mathbb{H}^\ast(X, DR(\mathcal{F}, \nabla)) $$ need not be an isomorphism. So the question is:

**Question 1**: Under what assumptions is this map an isomorphism? How do you prove it, and how do you put everything together to prove an analogue of Grothendieck's theorem?

**Question 2**: Does this story have a variant with compact support?