### Motivation

I was re-reading parts of Grothendieck-Murre, and these questions came up naturally.

The situation in chapter 9 is that $S'=Spec(A)$ where $A$ is a complete local noetherian ring of dimension $2$ with an algebraically closed residue field, and $D$ is some divisor in $A$. Then they take a "desingularization" of $S'$, $T'$, such that $T'$ maps isomorphically to $S'$ away from $D$, and the divisor that goes to $D$ is normal crossings. Call the divisor that maps to $D$ also $D$. They first show that $\pi_1^{(p)\,D}(T')$ is isomorphic to $\pi_1^{(p)}$ of the completion of $T'$ with respect to $D$ where this $\pi_1$ is the tame $\pi_1$ with respect to $D$. I assume this goes through even if $D$ is not normal crossings (if I'm wrong about this, I would really like to know).

Then they prove that $\pi_1^{(p)}(S'\setminus D)(=\pi_1^{(p)}(T'\setminus D))$ is isomorphic to $\pi_1^{(p),D}(T')$.

The question that arises is: why did they take this detour through this desingularization? Which step doesn't go through in the case that $D$ is not normal crossings? As I said, I think it's unlikely that it is the step that the tame fundamental groups are the same when completing, but I would like to know if I'm wrong.

### Question

Let $S'=Spec(A)$ where $A$ is a complete local noetherian ring of dimension $2$ with an algebraically closed residue field, and $D$ is some divisor in $A$. Could there exist a divisor, $D$ (necessarily not normal crossings), such that $\pi_1^{(p)}(S'\setminus D)$ is not isomorphic to $\pi_1^{(p),D}(S')$?