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Let $(M, \partial_{-}M, \partial_{+}M)$ be a decorated 3-cobordism, where $M$ be a (topological) 3-manifold and $\partial M=-\partial_{-}M \cup \partial_{+}M$. (decorated in a sense of Turaev, Quantum invariants of knots and 3-anifold, on page 159.) Suppose $M$ is homeomorphic to a cylinder over a torus $\Sigma \times I$, where $\Sigma$ is a torus $S^1 \times S^1$.

Let $f_{-}: \Sigma \to \partial_{-}M$ and $f_{+}: \Sigma \to \partial_{+}M$ be parametrizations of bottom and top boundaries respectively.

Consider the composition

$H_1(\Sigma; \mathbb{Z}) \to H_1(\partial_{-}M; \mathbb{Z}) \to H_1(\partial_{+}M; \mathbb{Z}) \to H_1(\Sigma; \mathbb{Z})$.

Here the first and the third ismorphism are induced by the parametrizations $f_{\pm}$ respectively and the second isomorphism (let's say $h$) is obtained by pushing loops in the bottom base of $M$ to the top base using the cylindrical structure on $M$.

Question

Is this cobordism $(M, \partial_{-}M, \partial_{+}M)$ (d-)homeomorphic to a cobordims $(\Sigma \times I, \Sigma \times 0, \Sigma \times 1)$, where the parametrization on the top is given by the identity of $\Sigma$ and the bottom is given by $f_{+}^{-1}hf_{-}$?

Please give me a proof. Thanks.

Edit

In this context, "decorated" can be almost ignored. What important here is that the homeomorphism of two cobordism is a homeomorphism of the two manifolds and if it is restricted on boundaries, it should commute with the given parametrizations.

The parametrization in this context means a degree 1 homeomorphism from a fixed torus to $\partial_{-}M$ and $\partial_{+}M$

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I don't have Turaev's book in front of me but it looks like either the result follows immediately or it might be slightly sensitive to the definition of "parametrization" together with the structure of the mapping class group of $\Sigma$. It would help if you could make your question more self-contained, including the conventions Turaev uses. –  Ryan Budney Feb 13 '12 at 8:07
    
I agree with what Ryan said. If it's just a question of whether the homeomorphism between the 3-manifolds commutes with the boundary parameterizations, then the answer to you question is clearly yes. Also, if anyone cares, this question is a continuation of mathoverflow.net/questions/87567 –  Kevin Walker Feb 13 '12 at 13:42
    
I think it should be trivial but I don't know how to prove it. Can you construct a homeomorphism? –  knot Feb 14 '12 at 2:02
    
I don't understand your question (in the above comment). You wrote "Suppose $M$ is homeomorphic to a cylinder...", so the homeomorphism between $M$ and $\Sigma\times I$ exists by assumption; there is no need to construct it. –  Kevin Walker Feb 14 '12 at 14:59
    
Yes, by assumption we have a homeomorphism $\varphi: M \to \Sigma \times I$. The parametrization of this cobordism is given by $f_{\pm}$ on top and bottom respectively. What I want to have is a homeomorphism from this cobordism to a cobordism $(\Sigma \times I, \Sigma \times 0, \Sigma \times 1 )$, whose parametrization is given by the identity on the top and $f_{+}^{-1}hf_{-}$ on the botton as in the question above. Does it make sence? –  knot Feb 14 '12 at 16:59
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