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Muestra que si f es la función que encaja a $S^{1}$ en la circunferencia que rodea al cilindro mayor, por la mitad, de la casa de Bing, entonces f es homotópica a una constante.

Show that if $\,f\,$ is the function embedding $\,S^1\,$ in the circumference around the main cylinder, at its half, of Bing's House, then $\,f\,$ is homotopic to a constant.


Here is the description of this space from page 4 of Hatcher's Algebraic Topology:

enter image description here

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  • $\begingroup$ ¿quien es Bing? $\endgroup$ – Eric O. Korman Apr 21 '13 at 2:29
  • $\begingroup$ Sería una buena idea que pusieras algo de trabajo propio que hayas hecho en esto. Yo no tengo ni idea, pero preguntas que no muestran un poco de esfuerzo hecho no son bien recibidas aquí muchas veces. $\endgroup$ – DonAntonio Apr 21 '13 at 2:30
  • $\begingroup$ @EricO.Korman, Chandler Bing, from "Friends"...:) Seriously: this seems to be a rather weird (for me) topological construction of something contractible but that looks...well, like a weird house.I never heard of it before. $\endgroup$ – DonAntonio Apr 21 '13 at 2:32
  • $\begingroup$ @DonAntonio so this question is asking about the inclusion of $S^1$ in a cone? $\endgroup$ – Eric O. Korman Apr 21 '13 at 2:41
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    $\begingroup$ No, Bing's House is the construction called "The House with Two Rooms" in Hatcher. This is relevant. $\endgroup$ – davidlowryduda Apr 21 '13 at 2:42
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In fact, Bing's House is deformation contractible to a point. It's easiest to see this by starting from a point and deforming it into Bing's House. A point is deformation-equivalent to a ball. Deform the ball into a cylinder. Now push into the cylinder one hole in the top and the bottom (but don't connect them!) and create two rooms. If you do this in the right way, you get Bing's House.

Now, in Spanish.

Es un hecho que la Casa de Bing es 'contráctil' (o contraible) a un punto. Ver esto es un poco dificíl, pero es más fácil si empezamos con un punto y lo deformamos a la forma de una casa. Un punto es deformable a un cilindro. Ya, se puede formar una impresión (depresión) en la parte de arriba, y otra en la parte de abajo, del cilindro (¡pero sin unirlas!). Y se puede extender las dos impresiones sin destruir las paredes separandolas. Al final, se puede crear la casa de Bing.

Por favor, corregir mi español.

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  • $\begingroup$ Your spanish is better than the one of many mexicans I know. Yet in mathematics sometimes little things can cause big differences. Point = punto, not punta which means "extreme of something" ( la punta del cuerno = a horn´s extreme , etc.). $\endgroup$ – DonAntonio Apr 21 '13 at 3:27
  • $\begingroup$ @Don: Ah, thanks quite a bit. Years and years ago, before I did any math, I lived in Mexico. Thank you for the corrections $\endgroup$ – davidlowryduda Apr 21 '13 at 4:10
  • $\begingroup$ Just tiny changes in the Spanish. It could just be the dialect of where I am from. $\endgroup$ – cactus314 May 18 '15 at 21:43

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