Although this question is in the same vein as my previous query,
it is nonetheless a different question.
Consider a compact, connected, smooth Riemann surface $X$.
Consider two line bundles now: $M=X\times\mathbb C$, the trivial line bundle, and a line bundle $N$ defined in the following way. Pick a point $x\in X$ and an open neighbourhood $U$ of $x$. Equip $U$ with a (continuous) coordinate function $z$ that vanishes only at $x$. The cover $\{U,V\}$, where $V=X\backslash\{x\}$, and a transition function $g_{UV}:(y,v)\mapsto(y,z(y)v)$, where $y$ is in $U\cap V$, are enough data to define $N$.
There are several ways to see that $M$ and $N$ are non-isomorphic as topological line bundles. For one, $M$ admits a global non-vanishing section while $N$ does not. In terms of characteristic classes, the first Chern classes $M$ and $N$ are different. Since $X$ is a Riemann surface, $H^2(X;\mathbb Z)$ is integral, and so we can rephrase this by saying that the degree of $M$ is $0$ and the degree of $N$ is $1$.
But how can I see directly, from first principles by writing down bundle maps, that $M$ and $N$ cannot be isomorphic? These are fairly simple bundles after all.
Idea:
Let's first refine the cover from $M$ so that it is the same as that of $N$. In other words, $M$ is the line bundle given by the cover $\{U,V\}$ and the transition $t_{UV}=(\mbox{id},1)$.
If $h:M\to N$ is an invertible bundle map, then it acts on $t_{UV}$ as $h_Vt_{UV}h_U^{-1}$, where $h_U$ is a map from $U\times\mathbb C$ to itself (acting as identity on $U$, and linearly and non-vanishing on $\mathbb C$), restricted to $U\cap V$. Likewise for $h_V$.
If I take $h_U$ to be $(y,v)\mapsto(y,z(y)v)$ and $h_V$ to be $(\mbox{id},1)$, then these transform $t_{UV}$ into the $g_{UV}$ originally specified for $N$.
The problem must be with the vanishing of $z$ at $y=x$. However, is it possible to use a function $Z$ that agrees with $z$ on $U\backslash\{x\}$ but which doesn't vanish at $x$, without assuming that they are analytic?