This is something from algebraic topology, as I understand (and the index suggests that retractions are covered mainly in the second part of the book). But, in fact, there was a fact about retractions (and that was the reason to look up the index) in the section on the quotient maps. A retraction (a continuous map that preserves all points in the image) is a quotient map.(a) Let $x\in Z-Y$ , let $y=r(x)$ . There are disjoint neighborhoods $U$ and $V$ of $x$ and $y$ . $V\cap Y$ is open in $Y$ and contains $y$ . $W=r^{-1}(V\cap Y)$ is open in $Z$ and contains $x$ and $y$ . If $z\in Y\cap W$ then $z=r(z)\in V$ and $z\notin U$ . Therefore, for every $z\in U\cap W$ , $z\notin Y$ .(b) The preimage of each point must be open and non-empty, and the union must be the whole space, which is connected.(c) For $x$ take $x/|x|$ . It is well-defined for all non-zero points, and continuous as a composition of continuous functions. Moreover, it preserves all points on the unit circle. As for the second question, it seems a little more trickier: whether $S^1$ is a retract of $\mathbb{R}^2$ . Suppose a continuous function $f$ maps $\mathbb{R}^2$ continuously onto the unit disc (not necessarily unit circle) and preserves all points on the circle. The restriction of the map onto the unit disc is a continuous map from the disc to the disc that preserves all the points on the circle. It seems that such a function must map some point to the origin (as well as to any other point in the disc). In other words, it must map the disc surjectively onto the disc. How to show this based on what we know? Another way is to show that a continuous map from the unit disc to itself must have a fixed point. Then we take the composition of $r$ with the rotation and the new continuous map does not have fixed points on the boundary, therefore, there is no continuous retraction onto the boundary. Yet, another way is just to try to prove that every circle centered at 0 maps to the entire boundary (under the retraction). This way when the circle vanishes to the origin, the origin must maps to all points on the boundary. This seems more promising. Let $r$ be the infimum of the set of radii such that every circle centered at 0 with the given radius or greater maps to the entire boundary. If $r>0$ then we show that $r$ is in the set and find $\epsilon$ such that $r-\epsilon$ is in the set. To find $\epsilon$ we, probably, need something like the uniform convergence.