If there is a continuous function from the closed unit disk to itself such that it is identity map on boundary, must it be onto?
General Topology – Continuous Function from Closed Unit Disk to Itself
algebraic-topologygeneral-topologyhomotopy-theory
algebraic-topologygeneral-topologyhomotopy-theory
If there is a continuous function from the closed unit disk to itself such that it is identity map on boundary, must it be onto?
Best Answer
Yes.
Suppose $f: D^2 \to D^2$ did not have $p \in D^2$ in the image, and such that $f$ restricts to the identity on the boundary. One may pick a homeomorphism $g: D^2 \to D^2$ that restricts to the identity on the boundary, with $g(p) = 0$, so $gf: D^2 \to D^2$ misses $0$.
Now compose with the map $D^2 \to S^1, x \mapsto x/\|x\|$. This defines a retraction $D^2 \to S^1$. But that's silly, as if there were such a retraction, the map $\pi_1(S^1) \to \pi_1(D^2)$ induced by the inclusion would be an injection, and it's not.