Let $X \subset \mathbb{R}^2$ be a triangle equipped with the topology induced by the euclidean topology on $\mathbb{R}^2$ and let $Y \subset X$ be the subset made of two sides of the triangle. I need to show that $Y$ is not a deformation retract of $X$.

This is an exercise from the first lectures of my course, so I'm not meant to use homotopy groups or that stuff. My tools are not much more than the definitions (and basic topology, of course).

The claim looks intuitive to me, but I'm stuck in trying to formalize it:

It's pretty easy to show that $Y$ is indeed a retract of $X$, so I need to show that, given a retraction $r:X\to Y$ and letting $i: Y \hookrightarrow X$ be the canonical inclusion, we have $ i \circ r \not \sim id_X$, where $id_X: X \to X$ is the identity function on X.

I though I could reason by absurd, assuming there is an homotopy $F:X \times [0,1] \to X$ between $i \circ r$ and $id_X$ and then get a contradiction with the continuity, i.e. showing that there is an open set of $X$ whose counter image through $F$ is not an open set of $X \times [0,1]$. However I'm having problems in finding such a set. Moreover it seems to me that I'm making the problem harder than it is…any help?

  • $\begingroup$ Can you argue that $i\circ r$ is null-homotopic, and the identity isn't, or does that use too much machinery? $\endgroup$ – Daniel Fischer Jul 18 '15 at 15:39
  • $\begingroup$ Yes but how can I show that the identity is not null-homotopic? $\endgroup$ – Manlio Jul 18 '15 at 15:45
  • $\begingroup$ If you know covering theory, you know that $\mathbb{R}$ is a (the universal) covering of the triangle, and a homotopy $\operatorname{id} \simeq \operatorname{const}$ would lift to the covering. That may be too much machinery, or not, depends on what you have to build on. $\endgroup$ – Daniel Fischer Jul 18 '15 at 15:49
  • $\begingroup$ Sorry, I fear this is too much machinery, covering theory will be sketched later on (according to the course schedule), way after homotopy groups. $\endgroup$ – Manlio Jul 19 '15 at 9:46
  • $\begingroup$ Another way to show that identity (on $S^1$) is not null homotopic. If it is you can use homotopy to extend $id$ to $D^2$ and ones you have a continuous map on $D^2$ identity on its border you have a contradiction as $S^1$ is not a retract of $D^2$ $\endgroup$ – Stan Jul 21 '15 at 15:24

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