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Let $X$ be a topological space and $V$ and $N$ are subspaces of $X$ such that $N$ deformation retracts onto $V$. I want to show that $X-V$ deformation retracts onto $X-N$. So i need to construct a retraction $r':X-V\longrightarrow X-N$ and show that there exists a homotopy between $i'\circ r'$ and $id'$ where $i':X-N\rightarrow X-V$ is the inclusion and $id'$ is the identity of $X-V$. So the main problem is to find $r'$, i know that we should use the retraction $r:N\longrightarrow V$ at some step but i don't see how to do it. thank you for your help!

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    You should also use the homotopy that witnesses the fact that $r$ is a deformation retract.2012-10-14
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    It would be very kind if you give me a hint on how to do that!!2012-10-14
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    Are you sure your statement is correct? Take $X$ a disk, $N$ its (closed) upper half and $V$ the equator. Then $N$ deformation retracts onto $V$, but $X-V$ isn't even connected.2012-10-14
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    Actually we need $V$ to be closed in $X$ and $N$ to be an open neighborhood of $V$ that deformation retracts on $V$ .2012-10-14
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    Still no good. If you fatten $N$ slightly in my previous example it still fits your conditions.2012-10-14
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    please see http://arxiv.org/pdf/1203.6097.pdf on page 4 proof of lemma 3.2 he uses this fact twice!!2012-10-14
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    He says "Let $A^1$ be the complement of an open tubular neighborhood of $QP(n−1)$ in $QP(n)$" few lines later he adds "$A^1$ is a deformation retract of $QP(n) − QP(n − 1)$"2012-10-14

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