Local homeomorphism : why isn't it $f:A\to B$ a local homeomorphism if for all $x\in A$, there is an open $U\ni a$ s.t. $f: U\to f(U)$ a homeo

Question

We say that $\varphi:A\to B$ is a local homeomorphism if for all $a\in A$ there is an open $U$ of $A$ and an open set $W$ of $B$ with $\varphi(a)\in W$ s.t. $$\varphi|_U: U\to W,$$ is a homeomorphism.

1) First of all, do we have to precise that $W\ni \varphi(a)$ ? Because if $\varphi|_U: U\to W$ is an homeomorphism, don't we have directly that $\varphi(a)\in W$ ?

2) Why we must take $W$ open in $B$ ? Don't such a definition work : for all open set $U\ni a$ the map $$\varphi : U\to \varphi(U)$$ is a homeomorphism. Why we want $\varphi(U)$ open in $B$ since it's open in $\varphi(U)$ already ?

Answer 1

1. We don't have to precise it, since it indirectly follows from the rest of the information: if $a \in U$ and $\varphi |_U : U \to W$, we have $\varphi(a) \in W$. But I (subjectively) find the original formulation easier to read, because when reading it for the first time and trying to imagine the setup, I immediately know that $W$ is a neighborhood of $\varphi(a)$, without having to do that (however trivial) reasoning.

2. Consider the mapping $\varphi : [0, 1] \to \mathbb{S}^1$ given by the standard formula

   $$\varphi(t) = ( \cos(2 \pi t), \sin(2 \pi t) ).$$

   The original definition of a local homeomorphism doesn't hold here but yours does. While the example doesn't decide which definition is 'better' (whatever it might mean), it certainly settles they are not equivalent.