MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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.

share|cite|improve this question
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
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.

share|cite|improve this answer
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

(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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.