We know that a sphere can be turned inside out. A nice video can be found here.
So now I wonder what happens to a genus $g$ surface $\Sigma_g$.
My Try: I basically want to use Smale Hirsch theory to calculate the path components of this space. Now the formal immersions correspond to maps from $\Sigma_g \to SO(3)$. But I am stuck at this point because I am not sure how one would calculate the homotopy classes of maps from $\Sigma_g$ to $SO(3)$. Any help is appreciated. Thanks
Recall that $SO(3) = \mathbb{RP}^3$.
If $X$ and $Y$ are CW complexes and $X$ has dimension $k$, then $[X, Y] = [X, Y^{(k+1)}]$ where $Y^{(k+1)}$ denotes the $(k+1)$-skeleton of $Y$; see this answer. In particular, if $Z$ is another space which has the same $(k+1)$-skeleton as $Y$, then $[X, Y] = [X, Z]$. In this case, $\Sigma_g$ has dimension two so $[\Sigma_g, SO(3)] = [\Sigma_g, \mathbb{RP}^3] = [\Sigma_g, Y]$ for any space $Y$ with $Y^{(3)} = \mathbb{RP}^3$ (as $\mathbb{RP}^3$ is three-dimensional, it is its own three-skeleton).
Note that $\mathbb{RP}^{\infty}$ has three-skeleton $\mathbb{RP}^3$. Moreover, it is a $K(\mathbb{Z}_2, 1)$. By the above reasoning we have
$$[\Sigma_g, SO(3)] = [\Sigma_g, \mathbb{RP}^3] = [\Sigma_g, \mathbb{RP}^{\infty}] = [\Sigma_g, K(\mathbb{Z}_2, 1)] = H^1(\Sigma_g; \mathbb{Z}_2) \cong \mathbb{Z}_2^{2g}.$$
Therefore there are $2^{2g}$ homotopy classes of maps $\Sigma_g \to SO(3)$.