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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.

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    $\begingroup$ Perhaps you want $M$ and $U$ to be compact, to rule out counterexamples like the one in Steve's answer? $\endgroup$ – HJRW Oct 17 '19 at 9:50
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    $\begingroup$ You might be seeking Van Kampen's theorem: en.wikipedia.org/wiki/Seifert%E2%80%93van_Kampen_theorem $\endgroup$ – Ian Agol Oct 18 '19 at 9:10
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    $\begingroup$ You should use stars *stars* in MathJax, not $\textit{math mode fakery}$ $\textit{math mode fakery}$, for italics. I have edited accordingly. $\endgroup$ – LSpice Oct 20 '19 at 2:09
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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.

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  • $\begingroup$ Dear HJRW, Thank you for your answer. Coming back to it now, after having done some reading about Van-Kampen and about graphs of groups (the latter I read from Hatcher), I realize I don't understand why Van-Kampen's theorem implies that $\pi_1(M)$ is presented as the fundamental group of the graph of groups. Now I should say that I do not come from geometric group theory and have little experience with its notions. 1/2 $\endgroup$ – Yaniv Ganor Dec 19 '19 at 13:40
  • $\begingroup$ The most general form of Van-Kampen I am comfortable with is Ronnie Brown's version with the groupoids and base points, (Although I am aware there is a more general version saying that $\Pi\left(\operatorname{hocolim} X_i\right)$ = $\operatorname{colim} \Pi\left(X_i\right)$ where $\Pi$ is the fundemental groupoid.) but still, I don't know how to use even that to show the claim about the presentation as fundamental group as graph of groups, or how to transition from knowing something about fundamental groupoids back to fundamental group. Is there an easy argument that I am missing? Thanks $\endgroup$ – Yaniv Ganor Dec 19 '19 at 13:41
  • $\begingroup$ @Blade, the above discussion is more or less equivalent to Theorem 1B.11 from Hatcher's Algebraic topology. The only difference is that Hatcher assumes that his vertex and edge spaces are apsherical. But you can make your vertex and edge spaces aspherical by attaching $n$ cells for $n\geq 3$. Since attaching such $n$-cells doesn't change the fundamental group, the result that you want follows. $\endgroup$ – HJRW Dec 20 '19 at 15:11
  • $\begingroup$ Or, to put it another way, Hatcher doesn't use the assumption that the vertex spaces are aspherical when proving that the inclusion-induced maps are injective. I hope that's some help. $\endgroup$ – HJRW Dec 20 '19 at 15:14
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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.

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    $\begingroup$ Or indeed one and none. This is a counterexample to the question as written (which needs some work), but not to reasonable perturbations of the question. For instance, if you remove small open discs instead of points, then it ceases to be a counterexample. $\endgroup$ – HJRW Oct 17 '19 at 9:48
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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?

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  • $\begingroup$ This is the proof of Britton's lemma. $\endgroup$ – HJRW Feb 24 '20 at 14:25
  • $\begingroup$ (It's nice to see it conveniently written out! :)) $\endgroup$ – HJRW Feb 24 '20 at 18:06
  • $\begingroup$ I am not too familiar with Britton's lemma, I saw the formulation in Wikipedia and books, but since the lemma seems purely algebraic I have to admit I don't see the analogy between the lemma and the above proof, can you elaborate? I am curious. Thank you very much for the help. $\endgroup$ – Yaniv Ganor Feb 25 '20 at 20:34
  • $\begingroup$ This method of proving Britton's lemma (more precisely, a cellular version of it) is sometimes called "t-corridors". It's well known to geometric group theorists, but I confess I'm amazed to discover that there aren't many introductory texts that include it. You could look at the proof of Britton's lemma in these notes of Short -- i2m.univ-amu.fr/~short/Papers/barcelona.pdf -- or alternatively at Exercise 7.2.4(ii) in this paper of Bridson: people.maths.ox.ac.uk/bridson/papers/bfs . $\endgroup$ – HJRW Mar 2 '20 at 15:27

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