Two lifts of a local homeomorphism

Question

Just learning about sheaves.

Suppose I have a sheaf $\mathscr{F}$ on a topological space. (The sheaf can take values in sets, let's say.) Let $E \overset{\pi}{\to}X$ be the etale space of this sheaf. Then $\pi$ is a local homeomorphism, but usually not a covering map.

A covering map $E' \overset{\pi'}{\to}X$, as we know, has the unique lifting property (if we assume things are nice), and to prove it, we let there be a map $f: U \to X$, where $U$ is connected, and we show that the points where two lifts of $f$ agree are

1. Open, (for which we use that given $x \in X$, there is a neighborhood $V$ of $x$ and a homeomorphism $V \to V' \subset E'$ which is a section of $\pi'$, and where $V'$ is open)
2. Closed, (for which we use that $E'$ is assumed Hausdorff, and for this it suffices to assume $X$ is Hausdorff by the covering map property)
3. Nonempty. (For which we assume that they agree at a single point.)

If we were to try this for $\pi$ instead of $\pi'$, definitely (2) would fail, because $E$ is usually not Hausdorff.

My Question: Would (1) fail also?

Clearly we don't have the same type of section of $\pi$ as we do in (1) above for $\pi'$. At first I thought that the definition of a germ would mean that if two lifts of $\pi$ agreed at a point, then they'd have to agree in a neighborhood of that point. But now I'm not so sure.

Answer 1

It is true : let $f:U\rightarrow X$ be any map and $g_1,g_2:U\rightarrow E$ be two lifts. Assume that $g_1(x)=g_2(x)=y$ and let $V$ be an neighborhood of $y$ such that $\pi:V\rightarrow \pi(V)$ is an homeomorphism. Then on the subset $U'=g_1^{-1}(V)\cap g_2^{-1}(V)\subset U$ the maps $g_1,g_2$ are given by $U'\overset{f}\rightarrow\pi(V)\overset{\pi^{-1}}\rightarrow V$ and so $g_1$ and $g_2$ coincide on $U'$ which is a neighborhood of $x$.

Using language from sheaf theory, assume that $U$ is an open subset of $X$, then $g_1,g_2$ are elements of $\mathcal{F}(U)$. Assume that their germs are equals at $x$, it exactly means that ${g_1}_{|U'}={g_2}_{|U'}$ for an open neighborhood $U'$ of $x$.

Answer 2

1) is also false. Consider the sheaf defined on $R$ which is the sheaf of sections of the trivial bundle $R\times R\rightarrow R$ a section is a continuous map defines by $f(x)=(x,h(x))$.

A section $f$ of the etale space here may be defined by a continuous map $g:U\rightarrow C(U,R)$ such that for $x\in U$, $f(x)=[g(x)]$ where $[g(x)]$ is the germ of $g$ at $x$. you can find two sections $f,f'$ such that $f(x)=f'(x)$ on $U$ if and only if $x=0$.