Fixed a topological space $S$, let $X\subset S\times \mathbb R^n$ be a subspace, with respect to the product topology, such that $(\{s\}\times \mathbb R^n)\cap X$ is a subspace of the vector space $\{s\}\times \mathbb R^n$, for any $s\in S$.
Fixed two naturals $m\leq n$, let $f:\mathbb R^n\to\mathbb R^m$ be a linear map, and define the map $F:X\to S\times \mathbb R^m$ as the restriction of $1_S\times f:S\times \mathbb R^n\to S\times \mathbb R^m$. So $F$ induces a linear map $F_s:(\{s\}\times\mathbb R^n)\cap X\to \{s\}\times \mathbb R^m$, for any $s\in S$.
Under the hypothesis that $F_s$ is bijective for all $s\in S$, is it true that $F$ is a homeomorphism? If not, are there some nice hypothesis we can add to guarantee that the question in italics has a positive answer?
For example, setting $m=n$ and $X=S\times \mathbb R^n$, then $F=1_S\times f$ and $F_s=f$ for all $s\in S$. So the assumption that the maps $F_s$ are bijective means that $f$ is bijective, hence it has a linear inverse $f^{-1}$, and $1_S\times f^{-1}$ is a continuous inverse for $F$. However I would be interested in some less trivial cases.
The motivation behind this question is that, when checking that a quasi vector bundle (embedded in a trivial vector bundle) is a vector bundle, one could find in the situation just depicted, as it happens in these two examples:
- The tangent bundle $TS^n\subset S^n\times \mathbb R^{n+1}$ on the $n$-dimensional sphere: for a point $x\in S^n$, define $U\subset S^n$ as the set of points $u\in S^n$ which are not orthogonal to $x$ (viewing $x$, $u$ as vectors in $ \mathbb R^{n+1}$). If $f:\mathbb R^{n+1}\to\mathbb R^n$ is the projection onto the subspace of $ \mathbb R^{n+1}$orthogonal to $x$, then $1_U\times f$ restricts to a homeomorphism $F:(U\times \mathbb R^{n+1})\cap TM\to U\times \mathbb R^n$. Hence we are in the situation above, with $S:=U$, $X:=(U\times \mathbb R^{n+1})\cap TM$, as $F_u$ is in fact bijective for all $u\in U$.
- The tautological line bundle $L\subset \mathbb RP^n\times \mathbb R^{n+1}$ on the $n$-dimensional real projective space: for a point $[x]\in \mathbb RP^n$, define $U\subset \mathbb RP^n$ as the set of points $[u]\in \mathbb RP^n$ such that $u$, $x$ are not orthogonal in $ \mathbb R^{n+1}$. Letting $f:\mathbb R^{n+1}\to\mathbb R$ be the projection onto the subspace spanned by $x$ in $\mathbb R^{n+1}$, $1_U\times f$ restricts to a homeomorphism $F:(U\times \mathbb R^{n+1})\cap L\to U\times \mathbb R$. Hence the situation is the same, with $S:=U$, $X:=(U\times \mathbb R^{n+1})\cap L$, as $F_{[u]}$ is in fact bijective for all $[u]\in U$.
Let $S = [0,1]$, $m = 1$, and $n = 2$. Take $$X = \left([0, 1) \times (\mathbb{R} \times \{0\})\right) \cup \left(\{1\} \times (\{0\} \times \mathbb{R})\right) \subset [0,1] \times \mathbb{R} \times \mathbb{R},$$ and let $f: \mathbb{R}^2 \to \mathbb{R}$ be the linear map $(x, y) \mapsto x + y$.
This satisfies all the conditions of the setup, but the map $F: X \to [0,1] \times \mathbb{R}$ is not a homeomorphism. In fact, $X$ is not even homeomorphic to $[0,1] \times \mathbb{R}$ via any map: the former space can be disconnected by removing one point while the latter cannot.
I don't think of any "nice hypotheses" that would guarantee the conclusion beyond demanding $X \to S$ be locally trivial: if $X$ is homeomorphic to a trivial bundle over $S$, then it is itself a vector bundle.