Continuous deformation of loop to point.

Question

Suppose I have a homotopy from a loop around the origin to a constant loop which is not the origin.

Prove that the origin is in the image of the homotopy.

Basically prove that if I deform a loop to a point, at some point in time it has to cross the origin.

I have tried to prove this but I get stuck trying to express mathematically that the loop can't break.

Answer 1

If the origin is not in the image, then consider your original loop as a loop in space $\mathbf{R}^2 \backslash(0,0)$.

Since your deformation avoids the origin, it can be viewed, again, as a deformation in the "punctured" space that deforms the loop encompassing the origin to the constant map.

But the loop in the constant map is not null-homotopic since the punctured space is equal to the circle, and the loop is nontrivial (homotopically speaking of course.)

[You may need to put more flesh on these!]

Answer 2

Assume your loop is $\theta:I \longrightarrow R^2$, and $F:I\times I \longrightarrow R^2$ is the homotopy taking it to the constant map.

Since $F(x,t)\ne 0$ for all values, then $$ G:I \times I \longrightarrow R^2 $$ $$G(x,t):=F(x,t)/\|F(x,t)\|$$ is well-defined and is a homotopy from loop $$\hat{\theta}:I \longrightarrow S^1 $$ $$\hat{\theta}(t)=\theta(t)/\| \theta(t)\|.$$ to the constant loop. It was meant for $\theta$ to be not null-homotopic, from which we can deduce that $\theta$ is not, either. But this is in contradiction with the existence of $G$. The existence of $G$ resulted from assumption that the origin is excluded. Thus, we proved that zero must be included in image of $F.$

Comment 1: I made one unproven claim: $\hat{\theta }$ is nontrivial.

Comment 2: We just proved (a simple adjustment needed) that if we have a deformation to constant map at origin, it has to "sweep" through ALL points in the interior of the loop.