Local behavior of sections of a local homeomorphism This is my first question on Math.stackexchange, so I hope somebody help me even if it results to be almost trivial.
My problem is about how to prove that two sections of a local homeomorphism coinciding on a point x, do also coincide over some neighbourhood of x.
Thanks for your attention.
 A: Suppose $f:X\to Y$ is a local homeomorphism, and that $\sigma_{1},\sigma_{2}:Y\to X$
are sections. This means that $\sigma_{i}(f(x))=x$ for every $x\in X$,
for both $i=1$ and $i=2$. Now, suppose $y_{0}\in Y$ is such that
$\sigma_{1}(y_{0})=\sigma_{2}(y_{0})$. We want to show that there
is a neighbourhood $V$ of $y_{0}$ such that $\sigma_{1}(p)=\sigma_{2}(p)$
for every $p\in V$.
Since $f$ is a local homeomorphism, there is a neighbourhood $W$
of $\sigma_{1}(y_{0})=\sigma_{2}(y_{0})$ such that $f(W)=:U$
is open and such that $f\restriction_{W}:W\to U$ is a homeomorphism.
For convenience, write $\varphi=f\restriction_{W}$. I claim
that there is a neighbourhood $V$ of $y_{0}$ such that $\sigma_{i}(p)=\varphi^{-1}(p)$
for every $p\in V$, for both $i=1$ and $i=2$; in particular, $\sigma_{1}(p)=\sigma_{2}(p)$
for $p\in V$.
Assume towards contradiction that there is no such neighbourhood $V$,
and fix some $i=1$ or $2$. This means that, for every
neighbourhood of $y_{0}$, I can find some $p$ in that neighbourhood
such that $\sigma_{i}(p)\neq\varphi^{-1}(p)$. Therefore, I can find
a converging sequence $(p_{n})$ such that $p_{n}\longrightarrow y_{0}$
and such that $\sigma_{i}(p_n)\neq\varphi^{-1}(p_{n})$ for every $n\in\mathbb{N}$.
(Alright, I'm assuming here $Y$ is first-countable; if this is a
big deal, just replace 'sequence' with 'net' and the same argument
should work). Since $\varphi:W\to U$ is injective, this means $\sigma_{i}(p_{n})$
is not in $W$ for every $n\in\mathbb{N}$. Moreover, since $\sigma_{i}$
is continuous, and $p_{n}\longrightarrow y_{0}$,
we get that $\sigma_{i}(p_{n})\longrightarrow\sigma_{i}(y_{0})$.
But $\sigma_{i}(y_{0})$ is in $W$ by definition, and $W$ is open,
and $\sigma_{i}(p_{n})\notin W$ for every $n\in\mathbb{N}$, so it's
impossible $\sigma_{i}(p_{n})\longrightarrow\sigma_{i}(y_{0})$. We
arrived at a contradiction.
I'm sure one can formulate the same argument in a neater way, but I hope this is better than nothing. Please tell me if something isn't clear.
A: I think more precisely what you want to show is the following: if $s_1$ and $s_2$ are sections of a local homeomorphism $\pi: E \to X$ between topological spaces (in this sense: https://en.wikipedia.org/wiki/Local_homeomorphism), and $s_1(x_0) = s_2(x_0)$ for some $x_0 \in X$, then there exists a neighborhood $x_0 \in U \subset X$ such that $s_1|_U = s_2|_U$. Here by a section we mean continuous maps $s_1, s_2: X \to E$ such that $\pi(s_1(x)) = \pi(s_2(x)) = x$ for all $x \in X$.
This follows by unraveling the definitions. Let $y_0 = s_1(x_0) = s_2(x_0)$. Since $\pi$ is a local homeomorphism, there are neighborhoods $y_0 \in V \subset E$ and $W \subset X$ such that $\pi|_V: V \to W$ is a homeomorphism (and hence has an inverse $(\pi|_V)^{-1}: W \to V$). By definition of sections, we have $x_0 \in W$ and by continuity, there is a neighborhood $x_0 \in U \subset W$ such that $s_1(U) \subset V$ and $s_2(U) \subset V$. Again by definitions, we have
\begin{align*}
&\pi(s_1(x)) = \pi(s_2(x)) = x \\
\implies{}&\pi|_V(s_1(x)) = \pi|_V(s_2(x)) \\
\implies{}&s_1(x) = ((\pi|_V)^{-1} \circ \pi|_V)(s_2(x)) \\
\implies{}&s_1(x) = s_2(x) \\
\end{align*}
for all $x \in U$. Hence $s_1|_U = s_2|_U$.
