I am trying to prove the following $\mathrm{Conn}^{\mathrm{reg}}(X) = \mathrm{Conn}(X) \cap \mathrm{Mod}_{rh}(\mathcal{D}_X)$. Here an integrable connection on a smooth algebraic variety $X$ is a $\mathcal{D}_X$ module which is locally free over $\mathcal{O}_X$ of finite rank. Regular integrable connections and regular holonomic modules are defined as in the book $\mathcal{D}$modules, perverse sheaves and representation theory. The inclusion of the lefthand side into the righthand side is easy because the composition factors of an integrable connection are integrable connections and regularity is closed under submodules, quotient modules and extensions. Where I got stuck is the other inclusion. Let $M$ be an integrable connection which is regular holonomic. Let $L_i$ a composition factor on $M$, then $L_i \simeq L(Y_i, N_i)$ where $N_i$ is a regular integrable connection on $Y_i$, and $Y_i$ is a locally closed smooth subvariety of $X$ such that the inclusion is affine. We notice that ($L_i$ is an integrable connection) we have $L(X,L_i) \simeq L_i \simeq L(Y_i, N_i)$. Therefore, by Theorem 3.4.2. on the same book, we have that $Y_i$ is dense (hence open being locally closed) and there exists a dense open subset $Y$ of $Y_i$ (which is, therefore, open dense in $X$) such that $L_i \vert_Y \simeq N_i \vert_Y$. Here is where I got stuck because I do not understand how I can conclude that the whole $L_i$ is a regular integrable connection, which is what I want to prove in order to prove that $M$ is a regular integrable connection.
$\begingroup$
$\endgroup$
This is actually just Theorem 6.1.6 (Curve Testing Criterion) combined with Defintion 5.3.2 (the definition of regular integrable connection).

$\begingroup$ Ok, I see how to prove the claim assuming that theorem. However, in the book the claim comes before that theorem. Is there a way to solve the problem without that theorem? $\endgroup$ – Federico Barbacovi Jul 5 '18 at 18:26

$\begingroup$ I haven't looked at the theorem in detail, but I think in the process of the proof, they actually reduce things to proving what you want. $\endgroup$ – Avi Steiner Jul 5 '18 at 18:27

$\begingroup$ I read the proof again and I think you are right. Thank you. $\endgroup$ – Federico Barbacovi Jul 5 '18 at 18:28