Does injectivity of $\pi_1(\partial U) \to \pi_1(M)$ imply injectivity of $\pi_1(U) \to \pi_1(M)$?

Question

Let $M$ be a smooth compact manifold of dimension $n$, and let $U$ be a smooth compact manifold with boundary, of the same dimension $n$, embedded in $M$.

The embedding induces maps on $\pi_1$.

If $\pi_1(\partial U) \to \pi_1(M)$ is injective, does this imply that $\pi_1(U) \to \pi_1(M)$ is injective?

If true, can you direct me to a reference or a short proof?

EDIT: I reformulated the question adding compactness of M and U to rule out the counterexample given in an answer.

Answer 1

The answer is 'yes' by Britton's lemma (see wikipedia and, more generally, Serre's book Trees and Scott and Wall's article 'Topological methods in group theory').

Since $M$ and $U$ are smooth and compact, $\partial U$ has a product neighbourhood $N(\partial I)\cong \partial U\times I$. Cutting along $\partial U$ realises $M$ as a graph of spaces, with vertices corresponding to $\pi_0(M-\partial U)$ and edge spaces corresponding to $\pi_0(\partial U)$. Van Kampen's theorem now asserts that the graph of spaces structure on $M$ induces a graph of groups structure on $\pi_1M$. Note that part of the definition of a graph of groups is that the edge maps should be injective -- this is where your hypothesis that $\pi_1(\partial U)$ injects comes in.

Finally, Britton's lemma (or, more precisely, its generalisation to the context of graphs of groups) implies that the natural maps from the vertex groups to the fundamental group inject, which is the fact you are asking for.

Answer 2

If $U$ is the closed disk, with two interior point removed, and $M$ is the closed disk with only one of those points removed, I think you have a counterexample.

Answer 3

I think I have found a somewhat elementary proof for the claim, and I would like to share it in case it is of interest for anyone, and also for some feedback.

The idea is to show the injectivity of $\pi_1(U)\to\pi_1(M)$ directly, by proving that every loop in $U$ that bounds a disc in $M$, bounds a disc contained in $U$.

Let $\gamma\colon S^1 \to U$ be a loop that is contractible in $M$, therefore there exists a map $u\colon D \to M$ such that $u\vert^{}_{\partial D}\equiv \gamma$, (here $D$ denotes the unit disk).

Without loss of generality we may assume that $\gamma$ is in the interior of $U$ (it can be homotoped inward away from the boundary), that $\gamma$ and $u$ are smooth (by Whitney's smooth approximation theorem) and that $u\pitchfork \partial U$ (by Thom's transversality theorem).

Consider the preimage $C=u^{-1}\left(\partial U\right)$, this is a compact one dimensional submanifold of $D$, hence it is a disjoint union of embedded closed curves, $C=\bigsqcup_j C_j$. We denote by $D_j$ the closed topological disc whose boundary is $C_j$.

Some of the curves $C_j$ may encompass others. We call a curve $C_j$ a maximal curve if it is not encompassed by any other component of $C$. More formally, $C_j$ is maximal if there exist no other component $C_k$ and a topological disc $D_k\subset D$, such that $C_j \subset D_k$ and $\partial D_k = C_k$.

Restricting $u$ to each of the maximal curves $C_j$ yields a loop $u\vert^{}_{C_j}$ contained in $\partial U$ and contractible in $M$ by $u\vert^{}_{D_j}$. Since $\pi_1(\partial U) \to \pi_1(M)$ is injective, the loop $u\vert^{}_{C_j}$ is contractible inside $\partial U$, and we thus can redefine $u$ on each of the disks that the maximal curves bound, such that $u$ now gives a contraction of $\gamma$ in $U$.

Does this make sense?