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Johansson's theorem states the following:

Given $f:M_1\rightarrow M_2$ (not a pair map) an homotopy equivalence between 3-manifolds with incompressible boundary.

Let $V_i$ be the components of the characteristic submanifolds meeting the boundary then we can homotope $f\simeq g$ so that $g$ is an homotopy equivalence on $V_i$ and is an homeomorphism on $\overline{M_1\setminus V_1}\rightarrow \overline{M_2\setminus V_2}$.

I also know about Waldhausen's theorem that if $f$ sends $\partial M_1$ to $\partial M_2$ homeomorphically then $f$ can be homotoped rel $\partial M_1$ to a homeomorphism.

My question is that if we know that $f\vert_{\partial M_1}$ is an embedding can we homotope $f:M_1\rightarrow M_2$ rel boundary to be an embedding?

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  • $\begingroup$ i don't understand the question. $\partial M_1$ and $\partial M_2$ are closed surfaces, so any embedding must necessarily be a homeomorphism, right? $\endgroup$ – ThiKu Jun 15 '16 at 14:18
  • $\begingroup$ I edited the question a little bit, the point is that $f$ might be mapping $\partial M_1$ into $\partial M_2$ so we cannot use waldhausen to homotope it to a homemomorphism $\endgroup$ – user93931 Jun 15 '16 at 14:29
  • $\begingroup$ You mean: "not mapping $\partial M_1$ into $\partial M_2$". But then a homotopy rel. $\partial M_1$ will preserve this property and thus can not yield a homeomorphism. $\endgroup$ – ThiKu Jun 15 '16 at 14:34
  • $\begingroup$ you are totally right I apologize $\endgroup$ – user93931 Jun 15 '16 at 14:46
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Yes, this is true, with the appropriate assumption that $M_1$ is irreducible (this is needed for Johansson and Waldhausen's statements), and let's say orientable.

One may reduce to Waldhausen's theorem if we know that $f_{|\partial M_1}$ is homotopic to a homeomorphism $f'_{|\partial M_1}:\partial M_1\to \partial M_2$, by homotopy extension.

Consider the image surface $f(\partial M_1) \subset M_2$ (assume this embedding is transverse, so doesn't meet $\partial M_2$). Since $\partial M_1$ is homologically trivial in $M_1$, $f(\partial M_1)$ is homologically trivial in $M_2$, hence bounds a submanifold $M \subset M_2$. Then $\partial M =f(\partial M_1)$ is homotopic to $\partial M_2$ iff $M_2-M$ is a product manifold (this follows from a result of Waldhausen, see Corollary 5.5 of

Friedhelm Waldhausen, MR 224099 On irreducible $3$-manifolds which are sufficiently large, Ann. of Math. (2) 87 (1968), 56--88.).

Take a homotopy inverse $g:M_2\to M_1$ to $f$. Then $g_{|M}$ is $\pi_1$-injective, and we may homotope $g$ so that $g_{|\partial M}\to \partial M_1$ is a homeomorphism. This map is also degree one, and $\pi_1$-injective, hence a homotopy equivalence. But this implies that $\pi_1M\to \pi_1 M_2$ is an isomorphism. In turn, this implies that $M_2-M$ is a product.

(I'm leaving out many details here, and I don't mean to give a proof by intimidation, so let me know if you would like to have any details on steps of this argument.)

As a reality check, the standard example of a homotopy equivalence which is not a homeomorphism is obtained by permuting the pages of a book of $I$-bundles. But the boundary of one will immerse in the other, and hence will not satisfy your hypothesis.

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  • $\begingroup$ Thanks for the reply. Why do we know that $f(\partial M_1)$ bounds a submanifold? I know we can find a complex representing it. $\endgroup$ – user93931 Jun 16 '16 at 7:28
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    $\begingroup$ One viewpoint is that $f(\partial M_1)$ is trivial in bordism homology, and hence bounds a submanifold. But this implicitly assumes one knows an identification of bordism homology with (say) singular homology. More directly, if $f(\partial M_1)$ does not bound a submanifold, then some component is non-separating in $M_2$. But this implies that this component is Poincaré dual to a 1-cycle. Transporting back to $M_1$, we get a contradiction, since the boundary components are not dual to 1-cycles, since they are separating. $\endgroup$ – Ian Agol Jun 16 '16 at 14:19
  • $\begingroup$ If I understand correctly we know that all of the $f(S_i)\in f(\partial M_1)$ are separating then we start splitting along each of them. Either we get a connected component whose boundary is $f(\partial M_1)$ or at least one of the $f(S_i)$ separates $f(\partial M_i)$. But the latter cannot happen since otherwise we would see the same in $M_1$, again by using that $f$ is an isomorphism in homology. Does this seem more or less correct? $\endgroup$ – user93931 Jun 16 '16 at 17:22

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