Let $f : B^2 \rightarrow \mathbb{R}^2$ a continuous map, is $B^2 \subset \operatorname{im}(f)$?

Consider a continuous map $$f : B^2 \rightarrow \mathbb{R}^2$$ such that $$f(S^1) \subset S^1$$ and $$deg(f_{|S^1}) \ne 0.$$

Prove that $$B^2 \subset \operatorname{im}(f).$$

[Note: Here, $$B^2$$ is the closed unit disk, and the degree of a function is defined as in this question]

I don't know how to prove the statement. In an informal way, I see that the winding number tells us that it's true but I don't know how the fundamental group and the degree encodes the information that the winding number gives.

The above idea comes from the following thread. But I don't know how to completely translate the solution to "purely" Algebraic Topology terms.

In my course, all the theory of the "Fundamental Group" and the "Degree" has been taught without any mention to the winding number, and everything I learned about the winding number is from a Complex Analysis course (i.e. with holomorphic functions.)

Thanks everyone!

Suppose that $$f$$ omits the point $$P$$ in the interior of the unit circle. There is a continuous map $$\Phi:R^2-\{P\}\to S^1$$ with the property that $$\Phi(Q)$$ is the point on the ray $$PQ$$ on $$S^1$$. Then $$\Phi$$ restricts to the identity on $$S^1$$. Then $$F=\Phi\circ f:B\to S^1$$ is continuous, $$F(S^1)\subset S^1$$ and the degree $$d$$ of $$F$$ on $$S^1$$ is the same as the degree of $$f$$ on $$S^1$$ (as they restrict to the same map from $$S^1$$ to itself).
But $$d$$ must be zero, as $$F$$ in effect gives a homotopy of $$F\mid_{S^1}$$ to a constant map, and this is a contradiction.
• Let $F:B^2\to X$ be a continuous map. Then $G:S^1\times I\to X$ defined by $G(z,t)=F(tz)$ is a homotopy from a constant map to $F\mid_{S^1}$. – Lord Shark the Unknown Apr 29 at 5:19