A curve $C$ is the image of some continuous, nonconstant map $\gamma:I\to\mathbb{R}^n$ from an interval to Euclidean $n$-space. Any such map of image $C$ is called a parameterization of $C$. We say that $\gamma$ is irreducible iff for all proper closed subintervals $[a,b]\subset I$ with $\gamma(a)=\gamma(b)$, we have $\gamma(I\backslash (a,b))\ne C$ and $\gamma([a,b])\ne C$.
The intuition behind an irreducible parameterization is that it doesn't contain "redundant overlaps" which do not contribute to the image of the curve and can thus be removed.
For example, one reducible parameterization of the unit circle $S^1$ is $f(t):=(\cos t,\sin t)$ defined on $[0,4\pi]$ which wraps around the circle twice.
My question: Is it true that every curve has an irreducible parameterization? If not, under what restrictions is it true?
A possible sketch of proof (for $I=[0,1]$):
A map $g:I\to X$ is called light if it is not constant on any interval.
It is shown in [1, Lemma 1.3.3], [2, Proposition 3.7] and [3] that, for every continuous surjection $g:I\to X$, there exists a continuous light surjection $\hat{g}:I\to X$ and a monotonically increasing surjection $m:I\to I$ such that $g=\hat{g}\circ m$. In order to apply this result, we can "set" the value of $\gamma$ on every maximal "reducible interval" to equal those at the endpoints of that interval, to obtain a new parameterization $g:I\to C$ of $C$. Then we can find a continuous light surjection $\hat{g} :I\to C$ satisfying the conditions above. It only remains to show that $\hat{g}$ is irreducible.
My (failed?) attempt:
I have a feeling that if we let $I=[0,1]$, this can be proved with Brouwer's reduction theorem by using a property $P(X)$ defined by, for example, "$X$ is a union of countably many disjoint (possibly degenerate) closed intervals $\bar{I}\subseteq [0,1]$ such that $\gamma(X)=C$ and that for each pair of adjacent intervals with endpoints $a\le b<c\le d$ we have $\gamma(b)=\gamma(c)$." Then we can find a nonempty closed subset $K$ of $[0,1]$ which has property $P$ irreducibly, provided that $P$ is inductive. We can kind of "glue" the intervals in $K$ together, in the sense that we construct a map from $[0,1]$ to $K$ which is strictly increasing and only skips within $K$ degenerate intervals and the right endpoint of each nondegenerate interval. Now we compose this map with $\gamma$ to obtain a new continuous map $g$, which will be an irreducible parameterization of $C$ because otherwise there would be a nonempty proper closed subset of $K$ for which $P$ still holds.
That said, I have no idea how to prove that the property $P$ is inductive. The definition says a property $P$ is said to be inductive provided that when each set in a monotone decreasing sequence $A_1\supset A_2\supset\cdots$ of compact sets has property $P$, so also does $\bigcap_1^\infty A_n$. Would it be possible that this intersection becomes no longer a union of countably many disjoint closed intervals?
Neither do I have an idea how to generally "glue countably many intervals" in $K$. Suppose that these intervals can be arranged into an infinite sequence $([a_n,b_n])_{n=1}^\infty$ such that $a_{n+1}>b_n$. Then we may calculate their total length $l=\sum_{n=1}^\infty(b_n-a_n)$ and define the value of $f(x)$ at $x=(\sum_{n=1}^{i-1}(b_n-a_n)+t)/l$ to be $a_i+t$ for all $i\ge1$ and $0\le t<b_i-a_i$. Finally we define $f(1)=\lim_{n\to\infty}b_n$ and obtain an irreducible parameterization $g=\gamma\circ f$.
For the case when $b_{n+1}<a_n$ the construction is similar, but what about the other cases? Can we always partition $K$ into finitely many subsets, each of which can be arranged into an infinite sequence in the same way?
Background: I made a conjecture that an irreducible (defined in a similar way) parameterization $\gamma:S^1\to\mathbb{R}^2$ of a convex closed curve $C$ has a good property, that is, for every line $L\subset\mathbb{R}^2$ the inverse image $\gamma^{-1}(C\cap L)$ either has cardinality $\le2$ or is a nondegenerate arc. I have no idea how to prove it either, but I naturally think that, in the first place, every convex curve should have an irreducible parameterization, hence this question.
Best Answer
Yes, every curve parametrized by $[0,1]$ has an irreducible parametrization, and your ideas for proving this are mostly correct.
You have a lot of good ideas mixed in with a few that won't quite fly, but luckily the correct ideas are enough. That is, we can still make your inductive property approach work.
The problems.
The problematic part of your outline is that you don't know that you can decompose into countably many disjoint closed intervals in the way you describe. That is, even when $n=1$ and you consider the Cantor function, this function is constant on a countable union of intervals, and what is left over is not a countable union of closed intervals - rather it is a Cantor set. For the same reason your property $P$ is not inductive, as the Cantor set arises as an intersection of nested finite unions of disjoint closed intervals.
The easy part.
Firstly, we should get the easy part of the proof out of the way. Let $t\in[0,1]$ be the minimum value (which exists by compactness) such that $\gamma([0,t])=C$, then let $s\in[0,t]$ be the maximum value such that $\gamma([s,t])=C$. Without loss of generality we will take $s=0$, $t=1$, as we may certainly restrict and reparametrize $\gamma$ to ensure this. Then $\gamma([a,b])\neq C$ for any proper closed interval $[a,b]\subset [0,1]$.
It can be verified easily throughout the rest of this proof that none of our subsequent reparametrizations change this fact, and so we leave it to the reader to keep us honest while we find a reparametrization $\gamma_0$ such that $\gamma_0([0,1]\backslash (a,b))\neq C$ for all proper subintervals such that $\gamma_0(a)=\gamma_0(b)$.
Defining $P$ correctly.
Now define $P(X)$ to be the property that $X\subseteq [0,1]$ is closed, satisfies $\gamma(X)=C$, and satisfies $\gamma(a)=\gamma(b)$ for each component $(a,b)$ of $[0,1]\backslash X$.
We claim $P$ is inductive. To see this, suppose $X_i\supseteq X_{i+1}$ is a decreasing sequence with this property. Then certainly the intersection $X=\cap_{i=1}^\infty X_i$ is closed and satisfies $\gamma(X)=C$ (the latter can be seen from the fact that for each $c\in C$ the sets $\gamma^{-1}(\{c\})\cap X_i$ are nonempty, and so give a nested sequence of nonempty compact sets).
It remains to show that $\gamma(a)=\gamma(b)$ for each complementary interval $(a,b)\subseteq [0,1]\backslash X$. To see this, we suppose $0<\epsilon<\frac{b-a}{2}$. Denote $U_i=[0,1]\backslash X_i$, so that $\bigcup_{i=1}^\infty U_i=[0,1]\backslash X$. The sets $U_i$ form an open cover of $[a+\epsilon,b-\epsilon]$, hence have a finite subcover, which by monotonicity can consist of a single $U_i$. Then if $(a',b')$ is the component of $[a+\epsilon,b-\epsilon]$ in $U_i$, then since $U_i\cap X=\emptyset$, we have $$[a+\epsilon,b-\epsilon]\subseteq (a',b') \subseteq (a,b),$$ so that $a'\in [a,a+\epsilon)$ and $b'\in (b-\epsilon,b]$. Since $\gamma(a')=\gamma(b')$, letting $\epsilon\to 0$ gives $a'\to a$ and $b'\to b$, so we have $\gamma(a)=\gamma(b)$ by continuity of $\gamma$.
Completing the proof.
Having established that $P$ is inductive, let $X$ have property $P$ irreducibly, as per Brouwer's Theorem.
Your idea now works perfectly. Define $\gamma'$ to equal $\gamma$ on $X$ and to equal the common value $\gamma(a)=\gamma(b)$ constantly on every complementary component $(a,b)\subseteq [0,1]\backslash X$. Observe that if $[a,b]\subset [0,1]$ is any proper subinterval with $\gamma'(a)=\gamma'(b)$, and $(a,b)\cap X\neq\emptyset$, then $$\gamma'([0,1]\backslash (a,b))=\gamma(X\backslash (a,b))\neq C,\tag{1}$$ since otherwise we could let $X'=X\backslash (a,b)$ and obtain a smaller set with property $P$.
Now let $\gamma_0$ be a light mapping such that $\gamma' = \gamma_0\circ m$ for some monotone increasing surjection $m\colon [0,1]\to [0,1]$.
To complete the proof we must show that $\gamma_0([0,1]\backslash (a,b))\neq C$ for any proper subinterval $[a,b]$ with $\gamma_0(a)=\gamma_0(b)$. To see this, observe first that if we let $(a',b')=m^{-1}((a,b))$, (note that the inverse image of an interval under a monotone map is another interval), then we must have $(a',b')\cap X\neq\emptyset$, as otherwise $\gamma'$ is constant on $(a',b')$, implying by lightness of $\gamma_0$ that $m$ is constant on $(a',b')$, so that $m((a',b'))\neq (a,b)$, contradicting surjectivity of $m$.
But then we have $\gamma'(a')=\gamma'(b')$ and so from (1) we obtain $$\gamma_0([0,1]\backslash (a,b))= \gamma_0(m([0,1]\backslash (a',b')))= \gamma'([0,1]\backslash (a',b'))\neq C.$$