X not simply connected and X-x contractible - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T22:46:46Z http://mathoverflow.net/feeds/question/34878 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34878/x-not-simply-connected-and-x-x-contractible X not simply connected and X-x contractible Manuel Rivera 2010-08-08T02:10:40Z 2010-08-08T15:40:20Z <p>Hello,</p> <p>I was wondering if there is a nice counterexample to the following question.</p> <p>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.</p> http://mathoverflow.net/questions/34878/x-not-simply-connected-and-x-x-contractible/34880#34880 Answer by Andreas Blass for X not simply connected and X-x contractible Andreas Blass 2010-08-08T02:20:09Z 2010-08-08T02:20:09Z <p>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.</p> http://mathoverflow.net/questions/34878/x-not-simply-connected-and-x-x-contractible/34932#34932 Answer by damiano for X not simply connected and X-x contractible damiano 2010-08-08T15:40:20Z 2010-08-08T15:40:20Z <p>(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.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p>