# Injectivity of homomorphism between homology groups of manifold and its boundary

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 $$\partial (M\setminus U_\epsilon(D))$$.

• 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 $j_*$. – Najib Idrissi Oct 11 '19 at 9:19
• @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. – H-H Oct 11 '19 at 18:07
• 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. – Najib Idrissi Oct 12 '19 at 8:10

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.