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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 $\endgroup$ – Blade 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 $\endgroup$ – Blade 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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