I have a triangle with identified edges shown below with an open disk cut out of the interior (picture given below). There are two parts to the question:

Let $X$ be the topological space constructed by gluing together three edges of a solid triangle and cutting out the center disk, and let $A \subset X$ be the "inner" boundary circle. enter image description here

  1. Is $X$ homotopy equivalent to $S^1$?
  2. Is $A$ a deformation retract of $X$?

My first attempt as $1.$ is the answer no because $X$ has fundamental group $\pi_1(X,x) \cong \langle a | \ a^3 = id \rangle,$ I think. On the other hand, I know that $\pi_1(S^1,y) \cong \mathbb{Z}$, and because fundamental group is invariant under homotopy equivalence, $X$ cannot be homotopy equivalent to $S^1$.

If $A$ is just a copy of $S^1$, then this would imply that $A$ is not a deformation retract of $X$, by $1.$ However, that seems to easy, and I'm guessing that the identification in the gluing for the construction of this topological space makes $A$ not just a copy of $S^1$. If this is the case and $A$ is a deformation retract of $X$, is there a good way to prove this or find the explicit homotopy?

  • $\begingroup$ Sorry, I think I´m missing something. Why is $a^3=id$? $\endgroup$ – svelaz Dec 31 '16 at 20:24
  • $\begingroup$ Now that I think about it, that doesn't seem right. If we call a path around $A$ $b$, then maybe Seifert van Kampen theorem tells us $pi_1(X,x) = \langle a, b| \ a^3 b = id \rangle$. Is that correct? $\endgroup$ – swygerts Dec 31 '16 at 20:29

Here's what my intuition tells me to do.

Since the three "$a$" edges are all identified with each other, its going to be hard to deformation retract "away from" those three edges "towards" the circle $A$ in a well-defined manner; after all, the points on those edges are identified 3-to-1 in the quotient space. So it seems unlikely that the circle $A$ could be a deformation retract.

On the other hand, it seems straightforward to deformation retract "away from" $A$ "towards" the quotient of the three "$a$" edges. In fact, once I get that idea, it is obvious how to define the deformation retraction: simply follow rays away from the center of the $A$ circle.

So, the whole space deformation retracts onto the quotient of the three "$a$" edges. That quotient is just a circle, with "$a$" representing a path that goes one time around the circle.

So the answer to question 1 is "yes", and furthermore the fundamental group is an infinite cyclic group generated by the path homotopy class of $a$, which we can write as $\langle a \rangle$.

Now we can turn to question 2. Under the deformation retraction to the $a$ circle, a path going one time around the $A$ circle has image going three times around the $a$ circle: the picture you provided let's us see that homotopy. So, if there were also a deformation retraction to the $A$ circle then the infinite cyclic group $\langle a \rangle$ would be generated by $a^3$, which is patently absurd.

So the answer to question 2 is "no".

Note that this is all done with deformation retractions. It's not necessary to apply Van Kampen's theorem.


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