continuous map of connected set is connected, example: Proving the connectedness of this set. I thought I would try to use this to prove connectedness in this set if possible:
$$\{(x,y)\mid 1<x^2+y^2<4\}$$
$f(x,y)=x^2+y^2$
So since $(1,4)$ is connected in $\mathbb R$ so it this set, as my function is continuous? Could I do the same with $\{(x,y,z)\mid 1<x^2+y^2+z^2<4\}$?
I realized this is pre-image, which doesnt have to be true. But is this function valid, or continuous:
$$f(x)=\{(u,v)\mid u^2+v^2=x^2, x \in (1,2)\}$$
 A: I would solve this by considering the "polar coordinates" map: $g(x, y) = (r(x, y), \theta(x, y))$. This is a continuous bijection with continuous inverse (on the area of interest), so in fact we only need to consider $(1, 4) \times [0, 2 \pi) \subset \mathbb{R}^{\geq 0} \times [0, 2 \pi)$.
This is connected: it's a product of connected sets.
A: As was stated in the comments, the preimage of a connected set need not be connected even if the mapping in question is continuous.
In this case I would most likely try to prove that the set you are considering is path-connected. Indeed if we look at the set you have described it is equivalent to
$$ \{ x \in \mathbb{R}^2 : 1 < |x| < 2\}$$
If you draw this set on paper you will notice it is simply a 2-dimensional donut! Next I would try and come up with a way to connect any two points in the donut. When choosing this path make sure that you know how to prove that each part of your path is a part of your set! 

 One such path between two points could consist of a shift toward the origin such that when you shift the point by a relevant angle we get to the second point!

