Proof degree of surjective circumference function

1 files changed, 41 insertions, 5 deletions

diff --git a/ts/n4.lyx b/ts/n4.lyx
index 47a0c21..52c7efa 100644
--- a/ts/n4.lyx
+++ b/ts/n4.lyx
@@ -1903,16 +1903,52 @@ Si
 status open
 
 \begin_layout Plain Layout
-\begin_inset Note Note
-status open
+Si 
+\begin_inset Formula $f$
+\end_inset
 
-\begin_layout Plain Layout
-Demostración.
-\end_layout
+ no es sobreyectiva, existe 
+\begin_inset Formula $z_{0}\in\mathbb{S}^{1}$
+\end_inset
+
+ tal que 
+\begin_inset Formula $f(\mathbb{S}^{1})\subseteq\mathbb{S}^{1}\setminus\{z_{0}\}\cong(0,1)$
+\end_inset
+
+ con el homomorfismo 
+\begin_inset Formula $h:(0,1)\to\mathbb{S}^{1}\setminus\{z_{0}\}$
+\end_inset
+
+ dado por 
+\begin_inset Formula $h(t):=e(t+\theta_{0})$
+\end_inset
 
+, donde 
+\begin_inset Formula $e(\theta_{0}):=z_{0}$
+\end_inset
+
+.
+ Por tanto, si 
+\begin_inset Formula $\alpha_{f}(s):=f(e(s))$
+\end_inset
+
+, existe un levantamiento de 
+\begin_inset Formula $\alpha_{f}$
 \end_inset
 
+ dado por 
+\begin_inset Formula $\tilde{\alpha}_{f}(s):=\theta_{0}+h^{-1}(\alpha_{f}(s))$
+\end_inset
+
+, pues 
+\begin_inset Formula $e(\tilde{\alpha}_{f}(s))=e(\theta_{0}+e|_{\mathbb{S}^{1}\setminus\{z_{0}\}}^{-1}(\alpha_{f}(s))-\theta_{0})=\alpha_{f}(s)$
+\end_inset
 
+, pero entonces 
+\begin_inset Formula $\tilde{\alpha}_{f}(1)-\tilde{\alpha}_{f}(0)=h^{-1}(\alpha_{f}(1))-h^{-1}(\alpha_{f}(0))\overset{\alpha_{f}(1)=\alpha_{f}(0)}{=}0$
+\end_inset
+
+.
 \end_layout
 
 \end_inset