Nodal curve in a smooth variety with injective map on fundamental groups 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?
 A: 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.
