When is a function not a local homeomorphism? my background is engineering and I am very new to the topology.
I think I got the concept of a local homeomorphism, but I cannot come up with a concrete example that is a continuous surjection but not a local homeomorphism.
For example, if I have a map $f:\mathbb{R}^n\to\mathbb{R}^m$ ($n>m$) that is a continuous surjection, in what case it is not a local homeomorphism? 
Also, is a projection map $f_{proj}:[0,1]\times[0,1]\to[0,1]$ where $(x,y)\mapsto x$ also a local homeomorphism? I thought it is since for every $(x,y)$ there is a neighborhood, e.g., $\{(\tilde x,\tilde y)|x-\epsilon<\tilde x<x+\epsilon, \tilde y = y\}$, which is homeomorphic to $\{\tilde x|x-\epsilon<\tilde x<x+\epsilon\}$. Is this correct?
EDIT:
Thank you for the comments and answers. I think I misunderstood the concept of the local homeomorphism. My question came from the covering map, $p:C\to X$, where the article says that every covering map is a local homeomorphism. Does it then mean that $C$ and $X$ have to be of the same dimensionality, or is it saying something different? Thank you in advance.
 A: For a function $f : X \to Y$ to be a local homeomorphism means that for every $x \in X$, there exists an open neighborhood $U$ of $x$ in $X$ and an open neighborhood $V$ of $f(x)$ in $Y$ such that $f : U \to V$ is a homeomorphism.
For $f : \mathbb{R}^n \to \mathbb{R}^m$, the fact that $f$ is a local homeomorphism implies $n=m$ is a deep result known as Invariance of domain theorem. It is a really important theorem, that basically states that the dimension is a topological notion.
For $f : [0,1]^2 \to [0,1]$, it is easier to show it, but still, you have to consider fundamental notions in topology. Basically, you have to look on the boundary and on the interior separatly. On a little open subset in the interior, removing a point on the left let it connected, but it is not the case on the right. 
In your last example, what you choose is not a neighborhood since it does not contain any open non-empty subset (because you fixed the $y$ coordinate).
Edit Careful not to misunderstand the definition of local homeomorphism. What you noticed in your example is that there exists a subset $E$ of $X$ for which the restriction and corestriction $f : E \to f(E)$ provides a homeomorphism. But this does not take account of the ambiant spaces $X$ and $Y$ of the definition here! A local homeomorphism tells something about $X$ and $Y$, not just about $f$.
More edit Take $p : X \to B$ a continuous function. With hands, saying it is a covering map means that $X$ locally looks like $B$, but is bigger, and that $p$ is the function htat makes it looks like $B$. More formally, it means that there is no difference between the local topology of $X$ and $B$, but there is a global difference that you can understand and control. There are many visual examples on wikipedia to understand the notion.
A: $f: \Bbb R^2 \to \Bbb R, f(x,y)=x$ is a continuous surjection, even an open map, but not a local homeomorphism: no neighbourhood of any point of the plane (open balls, say, or open squares) is homeomorphic to a set in $\Bbb R$ just by dimension considerations. 
If a metric space has local dimension $n$ everywhere, it also has global dimension $n$, by (classic but non-trivial) theorems in dimension theory. So in particular if $X$ is a covering space of $S^1$ say, it must be one-dimensional, like $S^1$. 
