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Let us fix a compact orientable 2 surface torus $\Sigma$. \Sigma=S^1 \times S^1$. We consider a cylinder $\Sigma \times I$ and a data $(\Sigma\times I, \Sigma\times 0, \Sigma\times 1, f_{0},f_{1})$. Here $f_{i}$, called parametrization, is a homeomorphism from $\Sigma$ to $\Sigma \times i$, for $i=0, 1$. Two such data are equivalent if there is a homeomorphism $\Sigma \times I$ to itself such that on the boundary it commutes with parametrizations.

Let us consider the composition

$H_1(\Sigma; \mathbb{Z}) \to H_1(\Sigma \times 0; \mathbb{Z}) \to H_1(\Sigma \times 1; \mathbb{Z}) \to H_1(\Sigma; \mathbb{Z})$.

Here the first and the third ismorphism are induced by the parametrizations $f_{i}$, $i=0, 1$ respectively and the second isomorphism (let's say $h$) is obtained by pushing loops in the bottom base of $\Sigma \times I$ to the top base using the cylindrical structure on $\Sigma \times I$.

Note that $h$ might not be the identity and might differ by each data.

The composition $f_{1}^{-1}hf_{0}$ of these gives an element of a mapping class group $M(\Sigma)$ of $\Sigma$.

Question: If two data induce the same element in $M(\Sigma)$, are these two data equivalent? Can you give a concrete homeomorphism $\Sigma \times I$ to itself that gives an equivalence?
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Let us fix a compact orientable 2 surface $\Sigma$. We consider a cylinder $\Sigma \times I$ and a data $(\Sigma\times I, \Sigma\times 0, \Sigma\times 1, f_{0},f_{1})$. Here $f_{i}$, called parametrization, is a homeomorphism from $\Sigma$ to $\Sigma \times i$, for $i=0, 1$. Two such data are equivalent if there is a homeomorphism $\Sigma \times I$ to itself such that on the boundary it commutes with parametrizations.

Let us consider the composition

$H_1(\Sigma; \mathbb{Z}) \to H_1(\Sigma \times 0; \mathbb{Z}) \to H_1(\Sigma \times 1; \mathbb{Z}) \to H_1(\Sigma; \mathbb{Z})$.

Here the first and the third ismorphism are induced by the parametrizations $f_{i}$, $i=0, 1$ respectively and the second isomorphism (let's say $h$) is obtained by pushing loops in the bottom base of $\Sigma \times I$ to the top base using the cylindrical structure on $\Sigma \times I$.

Note that $h$ might not be the identity and might differ by each data.

The composition $f_{1}^{-1}hf_{0}$ of these gives an element of a mapping class group $M(\Sigma)$ of $\Sigma$.

Question: If two data induce the same element in $M(\Sigma)$, are these two data equivalent? Can you give a concrete homeomorphism $\Sigma \times I$ to itself that gives an equivalence?
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Mapping class group and cylindrical structure

Let us fix a compact orientable 2 surface $\Sigma$. We consider a cylinder $\Sigma \times I$ and a data $(\Sigma\times I, \Sigma\times 0, \Sigma\times 1, f_{0},f_{1})$. Here $f_{i}$, called parametrization, is a homeomorphism from $\Sigma$ to $\Sigma \times i$, for $i=0, 1$. Two such data are equivalent if there is a homeomorphism $\Sigma \times I$ to itself such that on the boundary it commutes with parametrizations.

Let us consider the composition

$H_1(\Sigma; \mathbb{Z}) \to H_1(\Sigma \times 0; \mathbb{Z}) \to H_1(\Sigma \times 1; \mathbb{Z}) \to H_1(\Sigma; \mathbb{Z})$.

Here the first and the third ismorphism are induced by the parametrizations $f_{i}$, $i=0, 1$ respectively and the second isomorphism (let's say $h$) is obtained by pushing loops in the bottom base of $\Sigma \times I$ to the top base using the cylindrical structure on $\Sigma \times I$.

Note that $h$ might not be the identity and might differ by each data.

The composition $f_{1}^{-1}hf_{0}$ of these gives an element of a mapping class group $M(\Sigma)$ of $\Sigma$.

Question: If two data induce the same element in $M(\Sigma)$, are these two data equivalent?