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



Special retraction from a metric space onto an arc
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.




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**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.


