I was thinking about the spaces which are homotopy equivalent to $\mathbb{R}^2$ minus two points and I managed to confuse myself.

I know that this space is a deformation retract of wedge sum of two circles and the fundamental group of this space is the free group on two letters $\mathbb{Z} \ast \mathbb{Z}$. However, if we first remove one point from $\mathbb{R}^2$, we know that it is a deformation retract of a circle $S^1$. Then, we remove the second point from the circle $S^1$, we get a space that is homotopic to an open interval of $\mathbb{R}$. Since the fundamental group is a homotopy invariant, I would expect that these two spaces ($\mathbb{R}^2$ minus two points and an open interval) have the same fundamental group. But I know that they do not.

What am I missing?

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    $\begingroup$ It is not true because there are counterexamples! You have just found one. $\endgroup$ – Mariano Suárez-Álvarez Jan 24 '12 at 19:22

It does not work like that! Removing points does not commute with replacing spaces with homotopically equivalent ones (as you have just discovered :) )

A silly example: «being empty» is obviously a property of spaces which is homotopically invariant. Now, if we remove a point from $\mathbb R^2$ the resulting space is not empty; but $\mathbb R^2$ is homotopically equivalent to any one-point space $X=\{\star\}$, yet if we remove a point from $X$ we do get an empty space!

Later. That homotopy equivalence is an equivalence relation is quite irrelevant here, because the issue is not that but how that equivalence relation relates with the operation on removing points.

Another silly example: «Having most significant digit equal to $2$ » is a property of integers, and «being congruent modulo $2$» is an equivalence relation between integers. But $2000$ and $1000$ are equivalent under that relationship yet one of them has most significant digit equal to $2$ while the other does not.

  • $\begingroup$ But, isn't homotopy equivalence an equivalence relation? As you said, I've discovered removing two points at the same time and removing one by one are not equal but I still cannot see why this is true. $\endgroup$ – Karatug Ozan Bircan Jan 24 '12 at 19:23
  • $\begingroup$ @Karatug: Because another thing that is not invariant under homotopy is the property of "being different points". The point that is removed may become the same as a point that isn't removed! $\endgroup$ – Zhen Lin Jan 24 '12 at 19:27
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    $\begingroup$ @Karatug: Being an equivalence relation isn't relevant. For example, here is an equivalence relation on all spaces: any two spaces $X$ and $Y$ are equivalent. This is an equivalence relation on spaces yet a 1-point space is equivalent to a 2-point space. $\endgroup$ – Ryan Budney Jan 24 '12 at 19:32
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    $\begingroup$ @Karatug: $S^2$ minus a point is homeomorphic to $\mathbb R^2$, this is a much stronger statement than it being homotopy-equivalent to $\mathbb R^2$, in particular homeomorphism respects removal of subspaces. $\endgroup$ – Ryan Budney Jan 24 '12 at 19:33
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    $\begingroup$ Yes, exactly. Equivalence relations can be much weaker than homeomorphisms. $\endgroup$ – Ryan Budney Jan 24 '12 at 20:24

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