# Proving local injectivity of curve

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1. Feb 9, 2017

### MxwllsPersuasns

1. The problem statement, all variables and given/known data
Let γ : I → Rn be a regular smooth curve. Show that the map γ is locally injective, that is for all t0 ∈ I there is some ε > 0 so that γ is injective when restricted to (t0 − ε , t0 + ε ) ∩ I.

2. Relevant equations

3. The attempt at a solution

So I know a function (or a mapping) is injective when:
- {{∀a,b ∈ X}, ƒ(a) = ƒ(b) → a = b
And I get the concept here: We have an interval I and some regular smooth curve γ and when we restrict our interval to one point (t0) ± a very small number (ε) we can get a domain in which every element of the codomain is at most the image of one of the elements of the domain.

I imagine perhaps the notions of regularity and smoothness help in arguing the local injectivity of γ here but I'm just not seeing it and any help getting started on how to mathematically argue the local injectivity would be extremely appreciated!

2. Feb 10, 2017

### haruspex

I can think of an easy counterexample. Are you sure this is not a "prove or disprove" question?

3. Feb 11, 2017

### MxwllsPersuasns

The question specifically asks for you to show the local injectivity of the map γ. So is this something which is difficult to do? I'm guessing maybe by your response that it is. I mean, I think I understand conceptually how that works -- you know you can shrink the interval of a mapping such that eventually the mapping with be injective but I just don't know how to translate that to a mathematical argument.

However you are on to something as the second part of the question asks specifically "Give an example of a curve which is not (globally) injective"

4. Feb 11, 2017

### Staff: Mentor

What does regular in this context mean exactly? Does it rule out constant curves?

5. Feb 11, 2017

### MxwllsPersuasns

From my textbook - "A Parameterised curve is said to be Regular if it's velocity vector does not vanish anywhere"

6. Feb 11, 2017

### Staff: Mentor

So. And now for the second question. How does this property translate into terms of $\gamma (t)$?

7. Feb 11, 2017

### MxwllsPersuasns

Well so I would imagine that means that γ'(t) exists everywhere and thus it doesn't rule out the possibility of constant curves, since a constant curve is just one which the velocity vector equals a constant, right?

8. Feb 11, 2017

### Staff: Mentor

A constant curve would be $\gamma_0 : I \rightarrow \mathbb{R}^n$ defined by $\gamma_0 (t) \equiv v_0$ with a fixed point $v_0 \in \mathbb{R}^n \,.$ What is $\gamma'_0(t)$ then in this case? Is it injective? This would be a counterexample, if $\gamma$ wasn't required to be regular. However, it has to be regular.

Now $\gamma$ is smooth and $\gamma'(t) \neq \vec{0}$ for all $t \in I\,.$ Can you translate this into the limit definition of the differential?
For injectivity, as you have defined it above, you need to rule out the possibility, that for two points $a \neq b$ in a $\varepsilon -$neighborhood in $I$ holds $\gamma(a) = \gamma(b)$. Can you use the condition of regularity for this?

9. Feb 11, 2017

### MxwllsPersuasns

Okay well so if the curve itself equals a constant value v0 then it's derivative would be zero. Thus it wont change its value throughout the entirety of I. I think perhaps I was confusing that with constant-speed curve where the derivative would be constant and thus γ would indeed change its value - albeit in a constant fashion.

So γ0'(t) in this case would not be injective as we can see it takes every value of the interval I and maps it to 0.

For your next question would it just be something like γ'(t) = limΔt→0 {γ(t + Δt) - γ(t)}/{Δt} ≠ 0?

Would we take ε as the Δt in this scenario with the epsilon-neighborhoods??

Also I think I read somewhere (probably wikipedia) it is sufficient to state that the derivative never changes sign to prove injectivity within an interval. Is this true at all? Perhaps am I headed in the right direction?

Thanks so much for your help thus far fresh 42, can't tell you how much I appreciate it/have been enjoying the thought process.

10. Feb 11, 2017

### Staff: Mentor

Yes, that's the right direction. Just take a $t_0$ for which you want to show local injectivity and consider $\gamma'(t_0) \neq \vec{0}$
I would directly use the definition of the differentiation as a linear approximation, i.e. the limit which defines it, and try to show that the gap, that forces the velocity to be unequal to zero, also forces different function values of $\gamma (t)$ around $t_0$ , or the other way around, i.e. indirectly: if for all $\varepsilon > 0$ there is a $t_\varepsilon$ such that ... and so on.
Maybe it's more convenient to look at $||\, \gamma'(t_0) \,|| > 0$ but I haven't written down the equations, so maybe I'm wrong on this.

11. Feb 11, 2017

### MxwllsPersuasns

I'm terrible at methods of proving things (induction, counter-proofs, etc...) -- I'm actually enrolled in an intro to set theory and proof-writing course concurrently, whoops! -- but I believe I understand how to prove γ is indeed an injective function, on philosophical grounds:

So we see that the regularity condition imposes γ'(t) ≠ 0, ∀t ∈ I. Thus, γ(t) strictly increases or decreases as t traverses the interval I, but not both (as that would require: ∃t ∈ I such that γ'(t) = 0.) In other words, this means that once an element of the domain has been mapped to it's image in the codomain the next element's image will be strictly larger (if the derivative is positive) or strictly smaller (if the derivative is negative) thus eliminating the possibility that one element of the domain could share an image with another element of the domain.

Can anyone tell me if my reasoning is correct? I'm afraid though that using reasoning in this way will not be sufficient for my professor and I'll need to give a mathematical argument, a proper proof as it were. However as previously stated I'm awful at this so can anyone possibly help me translate this into a proof?

fresh 42 I know you said to use the limit def of the derivative as a linear approx and show that the difference γ(t + ε) - γ(t) which forces the velocity to be nonzero also forces γ(t) in an ε-neighborhood around t0 to take on different values but I'm still unsure how to do this? Maybe by saying that since γ'(t) ≠ 0 ∀t ∈ I which means γ'(t) will involve a strict inequality and so thus we'd have something like γ(t + ε) - γ(t) > 0 and so then we can just say that the epsilon neighborhood is what causes that offset from 0 and thus what spurns into motion all the rest about the strict inequality?

I'm sorry I just haven't had much experience with proofwriting and it was my understanding this course was more calculation/computation/verification based rather than proof based.

I appreciate all the help thus far! :)