Show that if \(f: S^m \into S^n\) and \(\exists U \subset S^m\) such that \(\restrictionof{f}{U} \cong f(U)\), then \(m=n\) and \(f\) is surjective.
Solution
1. The statement on the retained source sheet is false as written.
Take the standard equatorial embedding \(f:S^1\hookrightarrow S^2\). For every nonempty open arc \(U\subset S^1\), the restriction \(f|_U\) is a homeomorphism onto its image, but \(m=1\ne2=n\) and \(f\) is not surjective.
2. Even requiring \(f(U)\) to be open in \(S^n\) would not by itself imply surjectivity unless a global injectivity/local-homeomorphism hypothesis is added.
A map \(S^1\to S^1\) can map one open arc homeomorphically onto an open arc while collapsing or folding the rest of the circle into that same proper arc. Thus a local condition at one open set cannot force global surjectivity.
3. A standard valid correction is: if \(f:S^m\to S^n\) is continuous and injective, then \(m=n\) implies \(f\) is surjective; moreover an injective map between spheres can exist only when \(m\le n\), while a homeomorphism from a nonempty open subset of \(S^m\) onto an open subset of \(S^n\) forces \(m=n\) by invariance of dimension.
The dimension assertion is invariance of dimension. If \(m=n\) and \(f\) is injective, invariance of domain makes \(f(S^n)\) open in \(S^n\); compactness of the domain and Hausdorffness of the target make the image closed. Since \(S^n\) is connected and the image is nonempty, it is all of \(S^n\).