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1. Yes. 2. Yes. (I suppose that the surfaces are distinct if they are homeomorphic).

For 1, it is sufficient to check the definition of surface: that every point has a neigborhood homeomorphic to the disc. For interior points of the polygon, and for points on the sides, this is evident, and for the corners this is easy.

For 2, just recall classification of all possible compact surfaces up to homeomorphism. There is one integer invariant, the Euler characteristic, and the 2-valued invariant, orientability. The Euler characteristic is easily computed from your word: if you have 2n edges of your polygon, they will give $n$ edges after gluing, and suppose that you obtain $v$ vertices after gluing. Then the Euler characteristic is $1-n+v$ which is between $2-n$ and $2$. And orientability has two values. So you have at most $2n$ topologically different surfaces from strings of length $2n$.

And you see that for given length there are much more classes of words than $2n$.

1. Yes. 2. Yes. (I suppose that the surfaces are "the same" if they are homeomorphic).

For 1, it is sufficient to check the definition of surface: that every point has a neigborhood homeomorphic to the disc. For interior points of the polygon, and for points on the sides, this is evident, and for the corners this is easy.

For 2, just recall classification of all possible compact surfaces up to homeomorphism. There is one integer invariant, the Euler characteristic, and the 2-valued invariant, orientability. The Euler characteristic is easily computed from your word: if you have 2n edges of your polygon, they will give $n$ edges after gluing, and suppose that you obtain $v$ vertices after gluing. Then the Euler characteristic is $1-n+v$ which is between $2-n$ and $2$. And orientability has two values. So you have at most $2n$ topologically different surfaces from strings of length $2n$.

And you see that for given length there are much more classes of words than $2n$.

1. Yes. 2. Yes. (I suppose that the surfaces are distinct if they are homeomorphic).

For 1, it is sufficient to check the definition of surface: that every point has a neigborhood homeomorphic to the disc. For interior points of the polygon, and for points on the sides, this is evident, and for the corners this is easy.

For 2, just recall classification of all possible compact surfaces up to homeomorphism. There is one integer invariant, the Euler characteristic. The Euler characteristic is easily computed from your word: if you have 2n edges of your polygon, they will give $n$ edges after gluing, and suppose that you obtain $v$ vertices after gluing. Then the Euler characteristic is $1-n+v$ which is between $2-n$ and $2$.

And you see that for given length there are much more classes of words than possible values of the Euler characteristic.

1. Yes. 2. Yes. (I suppose that the surfaces are distinct if they are homeomorphic).

For 1, it is sufficient to check the definition of surface: that every point has a neigborhood homeomorphic to the disc. For interior points of the polygon, and for points on the sides, this is evident, and for the corners this is easy.

For 2, just recall classification of all possible compact surfaces up to homeomorphism. There is one integer invariant, the Euler characteristic, and the 2-valued invariant, orientability. The Euler characteristic is easily computed from your word: if you have 2n edges of your polygon, they will give $n$ edges after gluing, and suppose that you obtain $v$ vertices after gluing. Then the Euler characteristic is $1-n+v$ which is between $2-n$ and $2$. And orientability has two values. So you have at most $2n$ topologically different surfaces from strings of length $2n$.

And you see that for given length there are much more classes of words than $2n$.

1. Yes. 2. No. (I suppose that the surfaces are distinct if they are homeomorphic).

For 1, it is sufficient to check the definition of surface: that every point has a neigborhood homeomorphic to the disc. For interior points of the polygon, and for points on the sides, this is evident, and for the corners this is easy.

For 2, just recall classification of all possible compact surfaces up to homeomorphism. There is one integer invariant, the Euler characteristic. The Euler characteristic is easily computed from your word: if you have 2n edges of your polygon, they will give $n$ edges after gluing, and suppose that you obtain $v$ vertices after gluing. Then the Euler characteristic is $1-n+v$ which is between $2-n$ and $2$.

And you see that for given length there are much more classes of words than possible values of the Euler characteristic.

1. Yes. 2. Yes. (I suppose that the surfaces are distinct if they are homeomorphic).

For 1, it is sufficient to check the definition of surface: that every point has a neigborhood homeomorphic to the disc. For interior points of the polygon, and for points on the sides, this is evident, and for the corners this is easy.

For 2, just recall classification of all possible compact surfaces up to homeomorphism. There is one integer invariant, the Euler characteristic. The Euler characteristic is easily computed from your word: if you have 2n edges of your polygon, they will give $n$ edges after gluing, and suppose that you obtain $v$ vertices after gluing. Then the Euler characteristic is $1-n+v$ which is between $2-n$ and $2$.

And you see that for given length there are much more classes of words than possible values of the Euler characteristic.