X not simply connected and X-x contractible 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.
 A: (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.
A: 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.
