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Let $C$ be the nodal curve obtained by gluing together the points $0$ and $1$ of $\mathbb{A}^1_{\mathbb{C}}$. The topological fundamental group of $C$ is isomorphic to $\mathbb{Z}$.

One can find an immersion of $C$ in a smooth connected variety such that the map between the fundamental groups is non-trivial.

For example, take $\mathbb{A}^1_{\mathbb{C}}\times \mathbb{Z}/n\mathbb{Z}$ and glue $(1,k)$ with $(0,k+1)$ cyclically for every $k$. This produces a curve $C_n$ with a natural $\mathbb{Z}/n\mathbb{Z}$-action. The quotient of $C_n$ by the action of $\mathbb{Z}/n\mathbb{Z}$-action is isomorphic to $C$. Take an immersion of $C_n$ in a smooth variety $X_n$ with a compatible free $\mathbb{Z}/n\mathbb{Z}$-action. The quotient of $C_n\hookrightarrow X_n$ by the action of $\mathbb{Z}/n\mathbb{Z}$ is an immersion $C\hookrightarrow X$, where $X$ is a smooth variety. The image of $\pi_1(C)$ in $\pi_1(X)$, by construction, has $\mathbb{Z}/n\mathbb{Z}$ as a quotient. In particular it is non-trivial.

Does there exist a smooth connected variety $X$ and an immersion $C\hookrightarrow X$ such that the induced map $\pi_1(C)\rightarrow \pi_1(X)$ is injective?

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    $\begingroup$ I expect not. For example, the map on $H_1$ can't be injective because the weights on the mixed Hodge structures don't match. I realize that this comment may be a bit obscure, but I don't have the time write a detailed answer at the moment. $\endgroup$ Feb 11, 2017 at 22:39
  • $\begingroup$ Does a map of etale fundamental groups have closed image? That sounds like the limit of @DonuArapura's suggestion. If so, the discrete fundamental group of the nodal curve can't inject into the profinite fundamental group of of a normal variety. But what about varieties with fundamental groups that are not residually finite? Could they admit an interesting nodal curve? $\endgroup$ Feb 13, 2017 at 1:53
  • $\begingroup$ @Ben Wielend Every étale fundamental group is profinite by definition and continuous maps between profinite groups are always closed. Moreover the étale fundamental group of the nodal curve is $\hat{\mathbb{Z}}$ and not $\mathbb{Z}$. $\endgroup$ Feb 13, 2017 at 16:49
  • $\begingroup$ In SGA 3 they give an alternate definition of a pro-discrete etale fundamental group for which the fundamental group of a nodal curve is $\mathbb{Z}$. You could also look at Bhatt-Scholze's pro-etale stuff, where I believe the same is true. $\endgroup$
    – Joe Berner
    Feb 13, 2017 at 18:36
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    $\begingroup$ @MarcoD'Addezio The etale fundamental group is the group whose category of representations on sets or modules is the category of locally constant etale sheaves on the variety. If you restrict to actions on finite sets or modules over a finite field, then the group is profinite. But if you allow infinite sets or sheaves of $\mathbb Q$-vector spaces, then the nodal cubic has more local systems than $\mathbb G_m$. $\endgroup$ Feb 14, 2017 at 0:35

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Let me expand my comment slightly. By Deligne [Théorie de Hodge II, III], the homologies of complex algebraic varieties carry functorial mixed Hodge structures dual to the ones cohomology. Among other things, this means that the (co)homologies carry weight filtrations $W$ which are strictly preserved by induced maps. If $X$ is smooth then by construction, the possible weights of $H_1(X)$ are $-1,-2$, i.e. $Gr^W_iH_1(X)=0$ unless $i=-1,-2$. On the other hand, $H_1$ of your nodal curve $C$ has pure weight $0$. Therefore, for any map $C\to X$, the induced map on rational homology $H_1(C)\to H_1(X)$ must vanish. Using Morgan [The algebraic topology of smooth algebraic varieties], Malcev completions of $\pi_1$ carry mixed Hodge structures. (This is very roughly the inverse limit of the set of nilpotent quotients of $\pi_1$ upto torsion.) Using this, you can upgrade the argument to show that the map on Malcev completions $\hat{\pi}_1(C)\to \hat{\pi}_1(X)$ must vanish. This would settle your question in good cases.

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  • $\begingroup$ True, but there are interesting and natural Kahler groups with trivial Malcev completion, such as fundamental groups of irreducible Hermitian compact locally symmetric spaces of rank $>1$ (and even some complex hyperbolic surfaces). Can one get nonabelian Hodge theoretic restrictions in this case? $\endgroup$
    – Misha
    Feb 13, 2017 at 16:56
  • $\begingroup$ Yes, I agree that this isn't optimal. I think one can push this a bit further using Hain's relative Malcev completion or Mochizuki-Simpson's work. Perhaps I'll expand this later. $\endgroup$ Feb 13, 2017 at 17:04
  • $\begingroup$ For the nodal curve, $\pi_1 \cong H_1$. Your argument shows that the map on the level of $H_1$ must be zero, so that $\pi_1(C,c)$ has to map into the commutator subgroup of $\pi_1(X,c)$. Is this restriction stronger/weaker/orthogonal to Malcev completion? $\endgroup$
    – Joe Berner
    Feb 15, 2017 at 14:02
  • $\begingroup$ The kernel of $\pi_1\to \hat\pi_1$ lies in the kernel of $\pi_1\to H_1$, so Malcev gives more. $\endgroup$ Feb 15, 2017 at 15:52

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