Coincident point of map from mobius to $R^3$

Question

Assume $M$ is mobius. $f:M\rightarrow R^3$ is a continuous map , and map $\partial M$ to a circle on $xy$-plane ,and $z$ of $f(M)$ is non-negative.How to show that there must be different points $p,q\in intM$ such that $f(p)=f(q)$ ?

Answer 1

According to hint of Dan Rust.

Because the $z$ of $f(M)$ is non-negative. So ,we can glue half of sphere along the $\partial f(M)$ and the half sphere lie in the lower half space.

And there is not coincident point , so $f(M)$ like a half sphere . So, the result after gluing is homeomorphism to $S^2$ . This is equal to say that glue a disc along the boundary of Moebius will get $S^2$. But as we know , $S^2$ and disc are oriented , but Moebius is not . So we get contridiction (because it is equal to say that cutting a disc from $S^2$ , the result is not oriented).