Isomorphism of Vector Bundles Let $f:X \to Y$ be continuous map between topological spaces and $E$ a $n$-vector bundle over $Y$. Then we know that the pullback $f^*E$ is also a $n$-bundle over $B$ and a pullback in sense of category theory.
Let assume that we have following commutative square
$$
\require{AMScd}
\begin{CD}
F @>{g}  >> E  \\
@VVpV  @VVpV   \\
X @>{f}>> Y 
\end{CD}
$$
with another $n$-bundle $F$ and $g$ a bundle map. Since $f^*E$ is a pullback we obtain a continuous map $h:F \to f^*E$ of vector bundles.
My question is when is $h$ an isomorphism of vector bundles?
Does it suffice that $h$ induce a linear isomorphism $h_x: F_x \to f^*E_x$ on each fiber?
My considerations:
According to definition a morphism of vector bundles is an isomorphism iff
(1) it is a homeomorphism
(2)  fiberwise a linear isomorphism
$h$ is by construction continuous and bijective. (2) holds by assumption.
Is then $h$ already an isomorphism or are there some pathological conterexamples? The problem is that I can't find an argument that the inverse map is also continuous.
Or does it suffice to verify that the map is local homeomorphism using trivialisations?
 A: Claim: If $E\to X$, $E'\to X$ are two vector bundles over the same space $X$ and $f: E \to E'$ is a bundle map which is an isomorphism on each fiber, $f$ is an isomorphism, then $f$ is an isomorphism of bundles. 
Proof: We know that $f$ is bijective; to show $f$ is an isomorphism, we need to show that it is a homeomorphism. As this is a local property, we may reduce to the case when $E $ and $E'$ are trivial bundles, i.e. we have a map $f: X \times \mathbb{R}^n \to X \times \mathbb{R}^n$ which is an isomorphism on each fiber. Then $f$ is of the form $(x,v) \mapsto (x, g(x)v)$ for some continuous map $g: X \to GL_n(\mathbb{R})$. Then the inverse map $\eta: GL_n(\mathbb{R}) \to GL_n(\mathbb{R})$, $\eta(A) = A^{-1}$ is continuous, so the map $(x,v) \mapsto (x,\eta(g(x)) v)$ is continuous, and this is an inverse for $f$.

In general, a vector bundle map $X \times \mathbb{R}^n \to Y \times \mathbb{R}^k$ is of the form $(x,v) \mapsto (f(x), g(x)v)$ for $f: X \to Y$ and $g: X \to \mathrm{Hom}(\mathbb{R}^n, \mathbb{R}^k)$, and any map of vector bundles looks like this map locally. This can be used to advantage.
