Is $\iota_*:H_1(M-K) \to H_1(M)$ surjective if $K$ is an embedded loop which is null-homotopic?

Question

Let $M$ be a closed manifold with $\dim M \geq 2$. Let $c: S^1 \to M$ be an embedding (a priori topological, but it can be assumed to be smooth if necessary) which is null-homotopic, and let $K:=c(S^1)$. Is $\iota_*:H_1(M-K) \to H_1(M)$ surjective, where $\iota_*$ is the map induced in homology by the inclusion $M-K \hookrightarrow M$?

If necessary, we can also assume that $\dim M = 2$ (in this case, I would like to avoid the result mentioned here), but I would like a general situation if possible.

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This problem is motivated by a way of proving that a null-homotopic loop separates a closed surface in two connected components. The step missing in the argument is the above statement. Although this is the motivation, I am interested in the problem above and not this last one.

Answer 1

A way two see this if $M$ is a $2$-dimensional oriented manifold is to consider a fundamental domain of $M$ which is $2g$-gon and $c(S^1)$ inside the $2g$-gon, $\pi_1(M-c(S^1))$ is the free subgroup generated by $2g$ elements, since $M-c(S^1))$ is the disjoint union of a contractible disc and a surface which retract to the boundary of the $2g$-gon. this implies that $H_1(M-c(S^1))$ is $\mathbb{Z}^{2g}=H_1(M)$.

Answer 2

For orientable n-manifolds $M$ you can argue by duality. If $P \subset M$ is a compact polyhedron (not necessarily embededded as a subpolyhedron of $M$), then there is a natural isomorphism $\gamma : H_q(M,M-P) \to H^{n-q}(P)$. In your case this implies that $H_1(M) \to H_1(M,M-K)$ can be identified via $\gamma$ with $H^{n-1}(M) \to H^{n-1}(K)$, the latter homomorphism being induced by the inclusion $i : K \to M$. Since $i$ is inessential, $i_\ast = 0$. Hence the homomorphism $H_1(M) \to H_1(M,M-K)$ is zero which implies that $\iota_\ast$ is an epimorphism.

You can use the above duality theorem for arbitray compact $P \subset M$, but in the general case you have to use Cech cohomology.

For non-orientable manifolds you can work with $\mathbb{Z}_2$-coefficients which gives you at least a partial result.

Answer 3

You can prove this using tubular neighborhoods and Mayer-Vietoris:

Consider a tubular neighborhood $N$ of $K=c(S^1)$; so we assume $c$ is at least smooth. Then $N$ is homotopy equivalent to $S^1$ and thus $H_1(N)\cong\mathbb Z$, and a generator of this group is the loop $K$. Also, we have that $M\setminus N$ is homeomorphic to $M\setminus K$. We have by M-V(the reduced homology version) an exact sequence $$H_1(M\setminus N)\oplus H_1(N)\rightarrow H_1(M)\rightarrow \tilde H_0(\partial N).$$

However, as $K$ is connected, $\partial N$ is connected , and thus $\tilde H_0(\partial N)=0$, so the first map of the sequence is onto.

This means that any element of $H_1(M)$ is the sum of an element in $H_1(M\setminus N)$ and a multiple of $K$, thus $i_\ast:H_1(M\setminus K)\rightarrow H_1(M)$ is onto as we can just take a copy of $K$ in $M\setminus K$ that is homotopic to $K$ in $M$.