Regarding retractions of $X$ onto subspaces

Question

Let $A \subset X$ be a subspace of $X$. Recall that a retraction of $X$ onto $A$ is a continuous map $r: X \to A$ such that $r(a) = a$ for every $a \in A$.

Let $X = \bf R$ endowed with the standard topology, and let $A$ be the closed interval $[0,1]$ endowed with the subspace topology. Clearly, $[0,1]$ is open in $A$, but not in $X$. Am I right to say that, in this case, there is no retraction from $\bf R$ onto $[0,1]$, since $r^{-1}([0,1])$ is closed, and therefore $r$ is discontinuous?

Answer 1

You see in the answers and comments the existence of such a retraction.

I want to get to the point, which you have misunderstood:

Every topological space $Y$ is both, open and closed. And for EVERY function (in particular the continous ones) $f:X \to Y$, it holds that $f^{-1}$ is open and closed. This is because this preimage coincides with $X$. So in your example $r^{-1}([0,1]) = \mathbb R$, which is closed but also open, but not the same as $[0,1]$.

Hope it helps to clean up your confusion.

Answer 2

There exists a retraction $r:\Bbb R→I$ where $I=[0,1]$ $$r(x)=\begin{cases} 0,&\text{if }x\le0\\ x,&\text{if }0\le x\le1 \\ 1,&\text{if }x\ge1 \end{cases}$$ We could also write $r=\min(1,\max(0,x))$.