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.



X not simply connected and X-x contractible
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.


