# deformation retraction of the complement

Let $X$ be a topological space and $V$ and $N$ are subspaces of $X$. It is known that if $V$ is homotopy equivalent to $N$ then $X-V$ need not be homotopy equivalent to $X-N$, the Alexander horned sphere is an example. In the context when $N$ deformation retracts onto $V$ it is still true that $X-V$ needs not deformation retracts onto $X-N$ as illustrated by the example of $X$ is the disk and $V$ is the equator and $N$ is the upper half disk.

Now in a paper the author and for $Qp(n)$ the quaternion projective space he says "Let $A^1$ be the complement of an open tubular neighborhood of $QP(nā1)$ in $QP(n)$" then he says "$A^1$ is a deformation retract of $QP(n) ā QP(n ā 1)$"

Is there any additional property that let this happen in this case.. thank you for the clarification!!

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It might be more appropriate to remove the link to the paper, and just stick to the mathematical question. If it turns out there is an actual error in the paper, then the author could be contacted privately. – Steven Gubkin Oct 15 '12 at 16:22
There are some situations of reasonable generality where what you ask for is true, but as stated in your first paragraph, it's false. For example, if $X$ is a 3-ball, $V$ and $N$ an embedded interval or a point respectively. The complements do not have to be homotopy-equivalent. Your latter comment about the paper, though, that's fine. It's part of the tubular neighbourhood theorem. – Ryan Budney Oct 15 '12 at 16:28
What I mean to say is that statement in the paper is fine, and that's a fairly standard level of detail for that area. – Ryan Budney Oct 15 '12 at 16:28