Last time, we started proving the fact that every $g$-genus surface is topologically rigid where $g\ge 2$. The first two steps are given below:

1. Decompose $S_g$ into pair of pants using a finite collection $\mathcal{P}$ of a smoothly embedded simple closed curves, then homotope homotopy equivalence $f:S_g\to S_g$ to make it smooth as well as transverse to $\mathcal{P}$.

2. Thus, $f^{-1}(\mathcal{P})$ is a finite collection of smoothly embedded closed curves in $S_g$; some of these curves may bound disks in $S_g$, and others represent primitive elements of ${\pi }_1(S_g)$; next, we homotope $f$ to remove all disk-bounding components of $f^{-1}(\mathcal{P})$.



Today, we will start from here:

3. We first homotope $f$ to map each component of $f^{-1}(\mathcal{P})$ onto a component of $\mathcal{P}$ homeomorphically, this requires ${\pi }_1$-injectivity of $f$.

4. Therefore, any two components of $f^{-1}(\mathcal{P})$ co-bound an annulus if and only if they are inverse images of a single component of $\mathcal{P}$ as $f$ has homotopy left inverse. We further homotope $f$ to remove all annuli co-bounded by two components of $f^{-1}(\mathcal{P})$.

So, after all these homotopies, we have a one-to-one correspondence between $f^{-1}(\mathcal{P})$ and $\mathcal{P}$, in the sense that $f^{-1}(\mathcal{C})$ is a single circle mapped homomorphically onto $\mathcal{C}$ for each component $\mathcal{C}$ of $\mathcal{P}$.



If time permits, we will go through the next few steps:

5. Cut $S_g$ along $f^{-1}(\mathcal{P})$ and then show each complementary component is a pair of pants, and thus the whole problem reduces to proving the topological rigidity of pair of pants.