# weakly locally one-to-one?

Is there any standard name for this concept that is weaker than local one-to-one-ness?

In some open neighborhood of $x_0$ there is no point $x\ne x_0$ such that $f(x)=f(x_0)$.

Or, if you like: In some neighborhood of $x_0$, for every point $x$ in that neighborhood, $f(x)=f(x_0)$ only if $x=x_0$.

Might one simply say that "$f$ is weakly locally one-to-one at $x_0$"?

Trivial question, it might seem, if it's only about mathematics. Maybe it's about psychology of learning mathematics. I recently came across this error: If $f\;'(a)>0$, the $f$ is strictly increasing in some neighborhood of $a$. (Much less recently, I made that mistake myself, when, as an undergraduate, I was frustrated by my inability to write what I thought must be a simple $\varepsilon$-$\delta$ proof of that proposition.) Would people be less likely to make that mistake if somewhere in the recesses of their brain they subconsciously remembered hearing that "If $f\;'\ne 0$ at a point, then $f$ is weakly locally one-to-one at that point."?

• I've completely revised it. I realized it didn't say what I meant and came back to fix it, expecting a hundred downvotes and 50 indignant comments. Apparently I got here before they did. – Michael Hardy Nov 21 '11 at 18:30
• I was busy trying to figure out why weakly locally 1-1 wasn't equivalent to 1-1. :) – David Mitra Nov 21 '11 at 18:32
• Your last comment about $f'\ne0$ at a point is interesting, from a pedagogical standpoint (and in and of itself). I think your statement can be proven without MVT. Then you could ask if $f'\ne0$ at a point implies monotonicity near the point. Show that this is false, and then ask "what extra condition would be needed to get monotonicity"? – David Mitra Nov 21 '11 at 18:54
• To quote my discrete math professor: if you try to prove something and fail, and then find out this something is false then you did a good job not proving a false claim. – Asaf Karagila Nov 21 '11 at 22:24
• @DavidMitra : Certainly the following can be proved without MVT. If $f\;'(a)>0$, then in some neighborhood of $a$, for all $x$ in that neighborhood, if $x>a$ then $f(x)>f(a)$ and if $x<a$ then $f(x)<f(a)$. All you need to do is set $\varepsilon=|f(a)|$, then let $\delta>0$ be so small that for $x$ in the $\delta$-neighborhood of $a$, the difference quotient $(f(x)-f(a))/(x-a)$ is within $\varepsilon$ of $f\;'(a)$. That implies the difference quotient is positive, and the result easily follows. No MVT needed. – Michael Hardy Nov 22 '11 at 1:46

A map between topological spaces is called discrete if the preimage of every point is a discrete set, i.e., has no points of accumulation. [In this post all maps are continuous.] This property is weaker than being locally injective. As you observe, every differentiable function $f:(a,b)\to\mathbb R$ with nonvanishing derivative is discrete.
The discreteness property usually appears in tandem with openness (the image of every open set is open). For example, every discrete and open map between planar domains is the composition of a holomorphic function with a homeomorphism; this is Stoilov's theorem. In higher dimensions the equation (discrete and open)=(X)$\circ$(homeomorphism) remains unsolved for X. (One would like the elements of class X to satisfy some sort of Cauchy-Riemann equations.)
A fundamental result on discrete and open maps between $n$-dimensional topological manifolds is Chernavskii's theorem: the branch set (=the set of points where the map is not locally injective) has topological dimension $\le n-2$; in particular it does not separate the space.