# Is a finite CW complex minus a point still homotopy equivalent to a finite CW complex?

Let $X$ be a finite CW complex and $x_0$ a point in $X$.

My question is then just:

Is $X-\{x_0\}$ still homotopy equivalent to a finite CW complex?

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The answer is yes if $x_0$ lies in the interior of a top dimensional cell. –  John Klein Jun 11 '12 at 13:41
The answer is also yes if $X$ is a regular CW-complex (i.e. if all its attaching maps are homemorphisms). –  André Henriques Jun 11 '12 at 14:38

Let $X$ be the CW-complex obtained by attaching a 2-cell to the space $[-1,1]$ via the attaching map $S^1\cong [-1,1]/(-1\sim 1) \longrightarrow [-1,1]$ given by (the continuous extension of) $x\mapsto x\sin(1/x)$.
Then $X\setminus \{0\}$ is homotopy equivalent to an infinite wedge of $S^1$'s.
@Matin: I guess that Andr\'e thinks of $\mathrm{S}^1$ as the quotient of the interval $[-1,1]$ by its boundary. Its attaching map takes the same value at $-1$ and at $1$. –  Oblomov Jun 12 '12 at 8:11
@Andr\'e: Actually, are you sure that the resulting space is really a a infinite wedge of circles. It seems to me that it might even not be a CW complex but rather a ''sort of Hawaian earrings space. –  Oblomov Jun 12 '12 at 9:02
The Hawaiian earing space has a sequence of loops that converge to a constant loop. In the space I constructed, there is no sequence of loops converging to a constant loop, for the simple reason that the limit point has been removed. Here's another example where a similar phenomenon shows up: $\mathbb R^2\setminus \{1/2,1/3,1/4,...\}$ is h.e. to the Hawaiian earings, but $\mathbb R^2\setminus (\{0\}\cup\{1/2,1/3,1/4,...\})$ is homotopy equivalent to an infinite wedge of $S^1$s. –  André Henriques Jun 12 '12 at 9:07