Homotopy of maps from the circle to punctured plane Since I have no clue about topology, I am trying to understand the following both intuitively and by proof if it is simple enough.
Consider a map $f_{(a,b)}$ from the circle to the punctured plane $f:S^1\to \mathbb{R}^2\setminus\{0\}$ given by
$f_{(a,b)}(\phi)=(a+\cos\phi,b+\sin\phi)$
If $a^2+b^2\neq 1$, $f$ avoids the origin. All maps $f_{(a,b)}$ with $a^2+b^2<1$ are homotopic to one another, since to go from $f_{(a_0,b_0)}$ to $f_{(a_1,b_1)}$ a family of points $f_{(a_t,b_t)}$ can be found, connecting the points $(a_0,b_0)$ and $(a_1,b_1)$ within the open circle $a^2+b^2<1$.
The same holds for maps with $a^2+b^2>1$. Furthermore, going from a point where $a^2+b^2<1$, to a point where $a^2+b^2>1$ is not possible.
However, only the maps with $a^2+b^2>1$ are null-homotopic, why is that? How is it shown that this map is constant to a single point?
 A: The idea is that: let's first start with the most basic case: $\mathbb{R}^n$. In $\mathbb{R}^n$, any two maps are homotopic, which means that you can continuously deform one into another. If you have two functions $f$ and $g$, then you can define the homotopy: $[0,1]\times\mathbb{R}^n : (t,x) \mapsto tf(x) + (1-t)g(x)$.
For $t = 0$, this is $f$, and for $t = 1$, this is $g$. And everything stays continous in the process (provided that $f$ and $g$ are). To think of it intuitively: just think you're deforming the graph of $f$ into the graph of $g$, and that $t$ is your "time" axis for the transformation (look up videos or gifs for homotopies, it's better to visualize it before getting into the maths).
And the reason why we can do that in $\mathbb{R}^n$, is that for any time $t$, the deformation "stays" in $\mathbb{R}^n$. So for example, if you have a closed curve, you can just imagine contracting it down to one point.
However, now imagine what happens if you remove a point from $\mathbb{R}^n$. Then if a curve encloses that point you removed, then you will no longer be able to contract it down to a point:

The curve on the left can be shrunk down to a point, but the one on the right can't, because you removed the red point, which makes it impossible.
Now let's get to your problem. Visualizing a few examples on desmos, it looks like:

*

*if $a^2 + b^2 = 1$, then your curve is a circle passing through $0$:



*

*if $a^2 + b^2 < 1$, then your curve is a conic enclosing $0$:


*if $a^2 + b^2 > 1$, then your curve is a conic that does not enclose $0$:

Now, this is just a few pictures, and not a proper mathematical proof. But if you manage to prove these three claims, then this gets you what you want:

*

*in the first case, your curve "doesn't exist", because it passes through a "forbidden point" (the $0$ that you removed)

*in the second case, your curve encloses $0$, so you can't contract it down to a point.

*in the third case, you're fine, you can definitely shrink it down to a point without passing through $0$.

Note: on the pictures, all the curves are circles. I just said conic to be safe, maybe in some cases these could be other conics, or maybe not, it doesn't really matter. The point is whether or not these curves enclose $0$.
