memoire-m2

diff --git a/sections/introduction.tex b/sections/introduction.tex
@@ -45,32 +45,48 @@ example\dots
   = \beta\) with orientation. Even more so, if \(\alpha', \beta' \subset
   \Sigma\) are nonseparating curve such that each pair \((\alpha, \alpha'),
   (\beta, \beta')\) crosses only once, then we can choose \(\phi\) with
-  \(\phi(\alpha) = \beta'\) and \(\phi(\alpha') = \beta'\).
+  \(\phi(\alpha) = \beta\) and \(\phi(\alpha') = \beta'\).
 \end{lemma}
 
-% TODO: Prove the easy case too?
 \begin{proof}
-  Let \(\Sigma = \Sigma_{g, r}^p\) and consider the surface \(\Sigma_{\alpha
-  \alpha'}\) obtained by cutting \(\Sigma\) across \(\alpha\) and \(\alpha'\),
-  as in Figure~\ref{fig:change-of-coordinates}. Since \(\alpha\) and
-  \(\alpha'\) are nonseparating, this surface has genus \(g - 1\) and one
-  additional boundary component \(\delta \subset \partial \Sigma_{\alpha
-  \alpha'}\), so \(\Sigma_{\alpha \alpha'} \cong \Sigma_{g-1,r}^{p+1}\). The
-  boundary component \(\delta\) is naturally subdivided into the four arcs in
-  Figure~\ref{fig:change-of-coordinates}, each corresponding to one of the
-  curves \(\alpha\) and \(\alpha'\) in \(\Sigma\). By identifying the pairs of
-  arcs corresponding to the same curve we obtain the surface
+  Let \(\Sigma = \Sigma_{g, r}^p\) and consider the surface \(\Sigma_\alpha\)
+  obtained by cutting \(\Sigma\) across \(\alpha\), as in
+  Figure~\ref{fig:change-of-coordinates}. Since \(\alpha\) is nonseparating,
+  this surface has genus \(g - 1\) and two additional boundary component
+  \(\delta_1, \delta_2 \subset \partial \Sigma_\alpha\), so \(\Sigma_\alpha
+  \cong \Sigma_{g-1,r}^{p+2}\). By identifying \(\delta_1\) and \(\delta_2\)
+  we can see \(\Sigma\) as a quotient of \(\Sigma_\alpha\). Similarly,
+  \(\Sigma_\beta \cong \Sigma_{g-1, r}^{p+2}\) also has two additional boundary
+  components \(\delta_1', \delta_2' \subset \partial \Sigma_\beta\). Now by the
+  classification of surfaces we can find an orientation-preserving
+  homeomorphism \(\tilde\phi : \Sigma_\alpha \isoto \Sigma_\beta\). Even more
+  so, we can choose \(\tilde\phi\) taking \(\delta_i\) to \(\delta_i'\). The
+  homeomorphism \(\tilde\phi\) then descends to a self-homeomorphism \(\phi\)
+  the quotient surface \(\Sigma \cong \mfrac{\Sigma_\alpha}{\sim} \cong
+  \mfrac{\Sigma_\beta}{\sim}\) with \(\phi(\alpha) = \beta\).
+
+  As for the second part of the lemma, we consider the surface \(\Sigma_{\alpha
+  \alpha'}\) obtained by cutting \(\Sigma_\alpha\) across the arc determined by
+  \(\alpha'\). Since \(\alpha'\) is nonseparating, \(\Sigma_{\alpha \alpha'}
+  \cong \Sigma_{g-1, r}^{p+1}\) has one boundary component more than
+  \(\Sigma\), say \(\partial \Sigma_{\alpha \alpha'} = \delta \amalg \partial
+  \Sigma\). The boundary component \(\delta\) is naturally subdivided into the
+  four arcs in Figure~\ref{fig:change-of-coordinates}, each corresponding to
+  one of the curves \(\alpha\) and \(\alpha'\) in \(\Sigma\). By identifying
+  the pairs of arcs corresponding to the same curve we obtain the surface
   \(\mfrac{\Sigma_{\alpha \alpha'}}{\sim} \cong \Sigma\).
 
-  Similarly, \(\Sigma_{\beta \beta'} \cong \Sigma_{g-1, r}^{p+1}\) also has an
-  additional boundary component \(\delta' \subset \partial \Sigma_{\beta
-  \beta'}\) subdivided into four arcs. Now by the classification of surfaces we
-  can find an orientation-preserving homeomorphism \(\tilde\phi : \Sigma_{\alpha
-  \alpha'} \isoto \Sigma_{\beta \beta'}\). Even more so, we can choose
-  \(\tilde\phi\) taking each one of the arcs in \(\delta\) to the corresponding
-  arc in \(\delta'\). Now \(\tilde\phi\) descends to a self-homeomorphism
-  \(\phi\) the quotient surface \(\Sigma \cong\mfrac{\Sigma_{\alpha \alpha'}}{\sim}
-  \cong \mfrac{\Sigma_{\beta \beta'}}{\sim}\). Moreover, \(\phi\) is such that
+  Likewise, \(\Sigma_{\beta \beta'} \cong \Sigma_{g-1, r}^{p+1}\) also has a
+  boundary component \(\delta' \subset \partial \Sigma_{\beta \beta'}\)
+  subdivided into four arcs. By the classification of surfaces we can find an
+  orientation-preserving homeomorphism \(\tilde\phi : \Sigma_{\alpha \alpha'}
+  \isoto \Sigma_{\beta \beta'}\) taking each one of the arcs in \(\delta\) to
+  the corresponding arc in \(\delta'\). Hence \(\tilde\phi\) descends to a
+  self-homeomorphism \(\phi\) of the quotient \(\Sigma
+  \cong\mfrac{\Sigma_{\alpha \alpha'}}{\sim} \cong \mfrac{\Sigma_{\beta
+  \beta'}}{\sim}\). Finally, since \(\tilde\phi\) takes the arcs corresponding
+  to \(\alpha\) to the arcs corresponding to \(\beta\) and the arcs
+  corresponding to \(\alpha'\) to the arcs corresponding to \(\beta'\),
   \(\phi(\alpha) = \alpha'\) and \(\phi(\beta) = \beta'\), as desired.
 \end{proof}