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

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

2. Thus, $f^{-1}(\mathscr P)$ 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 $f^{-1}(\mathscr P)$.



Today, we will start from here:

3. We first homotope $f$ to map each component of $f^{-1}(\mathscr P)$ onto a component of $\mathscr P$ homeomorphically, this requires $\pi_1$-injectivity of $f$.

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

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



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

5. Cut $S_g$ along $f^{-1}(\mathscr P)$ 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.