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!
Complement of deformation retract
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0He says "Let $A^1$ be the complement of an open tubular neighborhood of $QP(nā1)$ in $QP(n)$" few lines later he adds "$A^1$ is a deformation retract of $QP(n) ā QP(n ā 1)$" ā 2012-10-14
1 Answers
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The proposition stated is false. But I had the same idea and came up with this:
The deformation retract can be viewed as a homotopy $R(t,x)$ such that $R(0,x)=id$ and $R(1,x)=r$. Note that fixing $x$ defines a collection of paths
$\gamma_x =R(t,x)$
from $x$ to $r(x)$. Suppose that the retract has the additional property, that there exists a set $\beta \in N$ (imagine the boundary) together with a continuous function $\phi : N -V \to \beta$, such that $\gamma_{\phi(x)}$ crosses $x$ exactly once.
Then $X-V$ deformation retracts onto $(X-N)\cup \beta$ by
$R^*(t,x)=\gamma_{\phi(x)} ((1-t_0)(1-t)) $
where $t_0$ is the number in $I$, such that $\gamma_{\phi(x)}(t_0)=x$. (Define $R^*(t,x)=id$ outside $N$).