# Retract (algebraic topology)

• Sep 7th 2009, 03:06 PM
Ecusphyl
Retract (algebraic topology)
Hi everyone, excuse me about my bad english but it isn't my natural language.
I have tried so much but I canīt solve the next problem:

Let $K$ be a closed orientable surface. Let $\alpha \subset K$ a loop such that $K - \alpha$ is connected. Show that $\alpha$ is retract of $K$.

I would like to know how to solve this.I hope sombody could help me. Thanks anyway.
• Dec 29th 2010, 02:49 PM
Rebesques
Choose two distinct points $x,y\in (K-a)^o$. The homogeneity lemma (cf Milnor's "Topology from the Differentiable Viewpoint") guarantees that there exists a
diffeomorphism of $K-a$ onto itself that maps one point to the other, and that this diffeomorphism is smoothly isotopic to the identity.
• Dec 29th 2010, 07:13 PM
xxp9
cut off one of the two generator circles of a torus, the resulting space is a cylinder which is connected, while a torus does not retract to a circle since its fundamental group is Z x Z. Did I understand your question correctly?
• Dec 29th 2010, 08:26 PM
TheArtofSymmetry
Quote:

Originally Posted by xxp9
cut off one of the two generator circles of a torus, the resulting space is a cylinder which is connected, while a torus does not retract to a circle since its fundamental group is Z x Z. Did I understand your question correctly?

A "deformation retract" and a "retract" are not the same thing. Your argument works if the question asked about a "deformation retract".

In other words, a torus does not deformation retract to a circle, but it retracts to a circle.