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I have question. The definition of Mapping class group is defined as follows:

Let $S$ be a closed orientable surface of genus $g\geq 2$. The mapping class group is defined as follows: $$M(S) = \frac{\mathrm{Homeo}^+(S)}{\mathrm{Homeo}^0(S)}.$$ Here, $\mathrm{Homeo}^+(S)$ is the set of orientation preserving homeomorphisms.

My question is: What is the definition of oreientation preserving homeomorphisms?

What do we mean by orientation preserving homeomorphism from $\mathbb{RP}^1$ to $\mathbb{RP}^1$?

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    $\mathbb{RP}^1$ and $S^1$ are homeomorphic. See https://math.stackexchange.com/q/4507224/349785 – Paul Frost Jan 10 '25 at 10:42
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    Do you know the definition of an orientation of a manifold? If so, it would be best to include the definition in your post, otherwise potential (and actual) answerers are left to guess what definition you have in mind. If not... well, go learn what an orientation is. – Lee Mosher Jan 10 '25 at 15:22

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If an $n$-manifold, $M$, is orientable, one definition of "an orientation" is a generator of $H^n(M)$.

Draw a circle as a representation of $\Bbb{RP}^1$ (where you understand that antipodes are identified). This is a $1$-manifold and $H^1(\Bbb{RP}^1) = \Bbb{Z}$. The two generators are "clockwise around this circle" and "counterclockwise around this circle" (understanding that you only get halfway around when the loop closes).

So orientation preserving homeomorphism means that the induced map on $H^n(M)$ is the identity (instead of the negation) map.

Eric Towers
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  • Right, but, a bit more precisely, a generator of $H^n_c(M)$, in the case when $M$ is connected and noncompact. (Of course, OP is asking about closed surfaces but one defines mapping class groups for noncompact surfaces as well.) – Moishe Kohan Jan 10 '25 at 18:15
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We will define what it means for $f:S\to S$, a homeomorphism, to be orientation preserving/reversing. You pick any point $p\in S$, since $S$ is a surface there is a homeomorphism $\mathbb{R}^2\to U$ where $U$ is an open subset of $S$ containing $p$ and the pre-image of $p$ is $(0,0)\in \mathbb{R}^2$. Likewise, there is another homeomorphism $V\to \mathbb{R}^2$ where $V$ is another open set of $f(p)\in S$ and this homeomorphism maps $f(p)\mapsto (0,0)$.

The composition will give you, $\mathbb{R}^2\setminus \{(0,0)\}\to \mathbb{R}^2\setminus\{(0,0)\}$. Now pick any loop. If this map changes the sign of the winding number then we say that $f:S\to S$ is orientation reversing.

Of course, you need to show that this definition is well-defined, regardless of the choices that you make, however, you assume that $S$ is orientable itself, and so the atlas can be chosen consistently to keep this definition well-defined.

There is another issue to address, that the definition does not depend on the initial point $p$ chosen. This comes down to the fact that $S$ is connected and so the orientation cannot suddenly change as you move through out the surface.

Nicolas Bourbaki
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