How does showing $\varphi \circ \iota_{\mathbb S^1}$ homotopic to $\mathrm{Id}_{\mathbb S^1}$ give us a contradiction to $f$ has no fixed points?

Question

I am trying to prove the $2$-dimensional version Brouwer's fixed point theorem.

If $f: \overline{\mathbb B^2} \to \overline{\mathbb B^2}$ is continuous and has no fixed point, and we define $\varphi: \overline{\mathbb B^2} \to \mathbb S^1$ by $\varphi(x)=\frac{x-f(x)}{|x-f(x)|}$, how does showing that $\varphi \circ \iota_{\mathbb S^1}$ is homotopic to $\mathrm{Id}_{\mathbb S^1}$ give us a contradiction?

Answer 1

Since $\overline{\mathbb B^2}$ is contractible, $\varphi$ is nullhomotopic, and hence so is $\varphi \circ \iota_{\mathbb S^1}$. But $\mathrm{Id}_{\mathbb S^1}$ is not nullhomotopic, so this is a contradiction.