Is mapping class group topological property ? This means that if $X$ and $Y$ are homeomorphic topological spaces then $MCG(X) \cong MCG(Y) $. is this true ? If no what is counterexample? When mapping class group is topological property? (Note that $MCG(X) $ is mapping class group of $X$ )
I use definition of mapping class group in the book " Braids, Links, and Mapping Class Groups. By Joan S. Birman " page 148:
Let $\mathrm{T}_{\mathrm{g}}$ denote a closed orientable surface of genus g; and let $z_{1}^{0}, \cdots, z_{n}^{0}$ denote $n$ fixed but arbitrarily chosen points on $T_{g} \cdot$ Recall that in Chapter 1 the symbol $\pi_{1} \mathrm{F}_{0, \mathrm{n}} \mathrm{T}_{\mathrm{g}}$ denoted the pure braid group on $\mathrm{T}_{\mathrm{g}}$ (with base point $\left.\left(z_{1}^{0}, \cdots, z_{n}^{0}\right)\right)$ and $\pi_{1} B_{0, n} \mathrm{T}_{\mathrm{g}}$ denoted the full braid group on $\mathrm{T}_{\mathrm{g}}$ (with the same base point). The following similar notation is meant to suggest relationships to be developed in this chapter: $F_{\mathrm{n}} \mathrm{T}_{\mathrm{g}}$ denotes the group of all orientation preserving homeomorphisms $h: \mathrm{T}_{\mathrm{g}} \rightarrow \mathrm{T}_{\mathrm{g}}$ such that, for each i, $h\left(\mathrm{z}_{\mathrm{i}}^{0}\right)=z_{\mathrm{i}}^{0}$ $B_{n} \mathrm{T}_{\mathrm{g}}$ denotes the group of all orientation preserving homeonorphisms $h: \mathrm{T}_{\mathrm{g}} \rightarrow \mathrm{T}_{\mathrm{g}}$ such that $h\left(\left\{z_{1}^{0}, \cdots, z_{n}^{0}\right\}\right)=\left\{z_{1}^{0}, \cdots, z_{n}^{0}\right\}$ These two groups are to be endowed with the compact-open topology.
$\pi_{0}\left(f_{n} T_{g^{\prime}}\right.$ id) denotes the group is called the pure mapping class group of $\mathrm{T}_{\mathrm{g}}$ $\pi_{0}\left(B_{n} T_{g},\right.$ id $)$ denotes the group of path components of $B_{n} T_{g}$ and is called the (full mapping class group of $\mathrm{T}_{\mathrm{g}}$. The notation $M(g, n)$ is also used for this group.