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Suppose $X$ is a metric space and $A$ is a subspace of $X$ homeomorphic to $[0,1]$ with its usual topology. Let $v$ an end point of $A$, that is $v$ does not disconnect $A$. Is there a retraction $r$ from $X$ onto $A$ such that $r^{-1}(v)=\{v\}$? Thanks

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    $\begingroup$ @BenoîtKloeckner There is a retraction, A is an Absolute Retract. $\endgroup$ Oct 23, 2014 at 21:38
  • $\begingroup$ This is not a bad question. (I am surprised that it got down-voted several times). $\endgroup$ Oct 23, 2014 at 22:03
  • $\begingroup$ @PedroPerez: sorry for my hasty (and now deleted) comment. $\endgroup$ Oct 24, 2014 at 15:23

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The answer is YES. Indeed, consider a metrics $\ \rho_A\ $ in A, topologically equivalent to the induced topology from $\ (X\ d),\ $ and such that $\ (A\ \rho_A)\ $ is isometric to the standard unit interval $\ [0;1]\ $ with the euclidean distance. Thus there is a function $\ f:[0;1]\rightarrow A,\ $ such that

$$ \forall_{t\in[0;1]}\quad\rho_A(v\ \,f(t))\ \ =\ \ t$$

Hausdorff theorem provides a metrics $\ \rho\ $ in $\ X,\ $ topologically equivalent to $\ d,\ $ and such that $ \rho|A\times A\ =\ \rho_A.\ $ Thus a required retraction $\ r: X\rightarrow A\ $ can be given as follows:

$$\forall_{p\in X}\quad r(p)\ :=\ f(\rho(v\ p))$$

Done, that's it.


REMARK   @Pedro carefully singled out the end-points of an arc. For other points of the arc the required retraction in general doesn't exist (which had to be obvious to Pedro). Indeed, consider a circle $\ S,\ $ and let a closed semicircle $\ A\ $ serve as the arc in $\ S.\ $ Then for every retraction $\ r : S\rightarrow A,\ $ and for every point $\ x\in A\ $ different from either of the end-points of $\ A,\ $ the inverse $\ r^{-1}(x)\ $ consists of more than one point.

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  • $\begingroup$ I do not know which Hausdorff theorem you are using. Could you give a reference? Thanks again. $\endgroup$ Oct 24, 2014 at 8:38
  • $\begingroup$ This is Hausdorff theorem about extending an equivalent metrics from a closed subset onto the whole space (Toruńczyk--much later of course--obtained a simple and elegant proof). This was mentioned in another thread on MO. $\endgroup$ Oct 24, 2014 at 10:38

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