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Hello,

I was wondering if there is a nice counterexample to the following question.

Suppose $X$ is a CW-complex which is not simply connected and there is a point $x\in X$ such that $X-x$ is contractible. Is $X$ homotopy equivalent to a wedge of circles? Maybe we do not even need the CW-complex condition.

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Take two 2-spheres. Glue them at the north pole ($x$) and the south pole to get a nonsimply connected space $X$ which becomes simply connected after removing $x$, –  Donu Arapura Aug 8 '10 at 2:18
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2 Answers 2

up vote 12 down vote accepted

Take a disconnected space $Y$ that isn't homotopically trivial, for example the disjoint union of two circles, and let $X$ be its suspension. Let $x$ be one of the two "vertices" of the suspension. $X$ isn't simply connected because there's a loop that starts at $x$, goes through one component of $Y$ to get to the other vertex, and returns through a different component of $Y$. If you remove $x$, what remains amounts to the cone on $Y$ (with a collar), so it's contractible (to the other vertex). And $X$ isn't homotopically equivalent to a wedge of circles because the non-trivial homotopy in $Y$ will produce non-trivial higher homotopy in the suspension.

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I think we hit upon the same example, but you give an explanation which is more helpful. –  Donu Arapura Aug 8 '10 at 2:34
    
What if we strengthen the condition such that $X$ is not simply connected and $X-x$ is contractible for all $x \in X$. Can we now classify $X$ up to homotopy equivalence? –  Manuel Rivera Aug 8 '10 at 13:19
    
I should have mentioned an even simpler example, starting with $Y$ being the disjoint union of a circle and a point. Then the suspension $X$ can be viewed as a 2-dimensional sphere together with one of its diameters. –  Andreas Blass Aug 8 '10 at 22:29
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(This should have been a comment to Andreas Blass' answer, but it did not fit there.) To answer the stronger question, asked in a comment to Andreas Blass' answer you can argue as follows in the case of a CW complex.

Suppose that $X$ is a CW complex, that $X$ is not simply connected, and that for any point $x$ in $X$ the space $X \setminus \{x\}$ is contractible, then $X$ is a circle.

If $X$ has cells in dimension at least three, then removing a point from the interior of such a cell does not change the 2-skeleton of the CW complex and hence does not affect the fundamental group (any homotopy between loops can be made to happen within the 2-skeleton). Since we are assuming that $X$ is not simply connected, but that the removal of any point makes the space contractible, it follows that $X$ cannot have cells of dimension three or more.

Similarly, removing a point in the interior of a cell of dimension two corresponds to removing a relation for the fundamental group of $X$. Again, since we are assuming that $X$ is not simply connected, the resulting space would have fundamental group surjecting to a non-trivial group and would therefore not be trivial. Therefore we deduce that $X$ has no cells of dimension two either.

We are left with $X$ having cells of dimension at most one. Thus $X$ is a wedge of circles and it is now easy to see that the stated condition implies that $X$ is in fact a single circle.

With similar arguments it seems that you can show also the following result. Suppose that $X$ is a CW complex such that for every point $x \in X$ the space $X \setminus \{x\}$ is contractible. Then either $X$ is itself contractible (e.g. $S^\infty$), or $X$ is homotopy equivalent (and maybe even homeomorphic) to a sphere.

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