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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. –  tomasz Oct 14 '12 at 18:08
    
It would be very kind if you give me a hint on how to do that!! –  palio Oct 14 '12 at 18:11
    
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. –  Miha Habič Oct 14 '12 at 18:14
    
Actually we need $V$ to be closed in $X$ and $N$ to be an open neighborhood of $V$ that deformation retracts on $V$ . –  palio Oct 14 '12 at 18:18
    
Still no good. If you fatten $N$ slightly in my previous example it still fits your conditions. –  Miha Habič Oct 14 '12 at 18:24
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1 Answer 1

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 $N$).

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