Let $X$ be an integral noetherian finite type scheme over an algebraically closed field $k$. Let $X'\to X$ be its normalization. Is the induced homomorphism of etale fundamental groups $\pi_1(X')\to\pi_1(X)$ injective?
If so, does this correspond to the fact that the functor Fet(X) $\to $ Fet(X') is fully faithful?
Both questions have negative answers. Begin with $\mathbb{P}^4$. Let $L\subset \mathbb{P}^4$, resp. $\Pi\subset \mathbb{P}^4$, be a line, resp. $2$-plane, and assume that $L$ and $\Pi$ are disjoint. Let $C\subset \Pi$ be a smooth projective curve of genus $g>0$. Let $f:C\to L$ be a finite morphism. Let $X$ be the union over all $t\in C$ of $\text{span}(t,f(t))\subset \mathbb{P}^4$. Then $X$ is simply connected. Yet $X'$ is a $\mathbb{P}^1$-bundle over $C$, hence it has fundamental group equal to the fundamental group of $C$.
To see that $X$ is simply connected, observe that both $L$ and every $\text{span}(t,f(t))$ are connected and simply connected. Thus, for every finite, étale morphism $u:Y\to X$, the curve $u^{-1}(L)$ is a disjoint union of copies of $L$. Fix one such copy, say $L_i$. For every $t\in C$, the curve $u^{-1}(\text{span}(t,f(t)))$ is also a disjoint union of copies of $\text{span}(t,f(t))$. There is precisely one such copy that intersects $L_i$. The union over all $t\in C$ of this copy of $\text{span}(t,f(t))$ is a copy of $X.$
Edit. An "official" reference for the argument in the second paragraph is Corollaire IX.6.11 of the following.
MR0217088 (36 #179b)
Grothendieck, Alexander
Revêtements étales et groupe fondamental. Fasc. II: Exposés 6, 8 à 11.
Séminaire de Géométrie Algébrique, 1960/61. Troisième édition, corrigée
Institut des Hautes Études Scientifiques, Paris 1963 i+163 pp.
There is a unique morphism $F:X\to L$ whose restriction to $X\setminus L$ is the composition of projection to $C$ followed by $f$. This morphism is projective and surjective. The fibers are connected and étale simply connected (a union of concurrent lines). Thus, by the Corollaire, $X$ is also étale simply connected.
