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I asked similar question before, after some modification, I have a new question. Suppose $M^{n\geq 4}$ is a connected compact smooth manifold with connected nonempty boundary. Suppose $i_*: H_1(\partial M)\rightarrow H_1(M)$ is not injective, for simplicity, assume that $\dim (\ker(i_*))=1$, that means, there is 1-cycle (choose a loop $C$) in $\partial M$ which is nontrivial in $\partial M$ but trivial in $M$, assume it is boundary of 2-dimensional embedded submanifold $D$ (let's say $D$ is a 2-disc) in $M$, now denote the small neighbourhood of $D$ in $M$ by $U_\epsilon(D)$ ($\epsilon$ can be arbitrarily small), now I am wondering if new homomorphism $$j_*: H_1(\partial(M\setminus U_\epsilon(D)))\rightarrow H_1(M\setminus U_\epsilon(D))$$ is injective now?

What I have tried: suppose $n=4$, the neighbourhood of $C$ in $\partial M$ is a solid torus, and any circle in the kernel of $i_*$ is homologous to $C$, thus homologous to a circle on the boundary of the solid torus, which retracts along the new boundary $\partial (M\setminus U_\epsilon(D))$.

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  • $\begingroup$ I don't think so, consider $M = S^2 \times D^2$, the boundary is $S^2 \times S^1$ which has a nontrivial loop (e.g. by fixing a point $x \in S^2$ and looking at $\{x\} \times S^1$) that becomes trivial in $M$. But even if you remove a neighborhood of $\{x\} \times D^2$, you can just fix another $y \in S^2$ (sufficiently far away) and $\{y\} \times S^1$ is in the kernel of $j_*$. $\endgroup$ – Najib Idrissi Oct 11 at 9:19
  • $\begingroup$ @NajibIdrissi, thanks for answering, what do you mean by "sufficiently far away", I think now matter where $y$ is, $\{y\}\times S^1$ is homologous to $\{x\}\times S^1$, thus is trivial in the new boundary. $\endgroup$ – H-H Oct 11 at 18:07
  • $\begingroup$ It's difficult to visualize for $n=4$ but consider the solid torus $M = S^1 \times D^2$ instead. If you remove a neighborhood of $\{x\} \times D^2$, you're left with $M \setminus U_\epsilon(D) = \mathbb{R} \times D^2$, whose boundary is $\mathbb{R} \times S^1$ which has a nontrivial $1$-cycle. $\endgroup$ – Najib Idrissi Oct 12 at 8:10
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I am not sure whether this is the kind of thing you want to know, but here is a procedure to make $\partial M \rightarrow M$ an isomorphism on $\pi_1$ if $\text{dim}(M) \geq 5$.

As $M$ is compact, $\pi_1 M$ is finitely presented. Given one of the finitely many generators of $\pi_1 M$, represent it by an embedded circle based in $\partial M$. Excising a tubular neighborhood of each such circle corresponing to a generator, yields a new manifold $M'$ and a map $\partial M' \rightarrow M'$ which is surjective on $\pi_1$.

There is a lemma (I think it is due to Siebenmann) that states that a surjective homomorphism of finitely presented groups has finitely normally generated kernel. Hence, we find finitely many disjointly embedded circles in $\partial M'$ that normally generate the kernel of $\pi_1(\partial M') \rightarrow \pi_1(M')$. Since $\text{dim}(M') \geq 5$, these circles bound embedded disks in $M'$. If we excise tubular neighborhoods of these, we obtain the desired manifold whose inclusion of the boundary is a $\pi_1$-isomorphism.

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