Cut a homotopy in two via a "frontier"

Question

Consider a space $G$ obtained by glueing two disjoint cobordisms (the fact that they are might be irrelevant, assume they are topological spaces at first) $L$ and $R$ on a common boundary $C$. (Another possible setting is glueing disjoint topological spaces on a point)

Now consider a continuous function $f$ from the square $[0,1] \times [0,1]$ to $G$ such that $f(0,-) \in L$ and $f(1,-) \in R$, that is on the left side of the square the function is mapped inside $L$ and to $R$ on the right side.

Now, on every horizontal slice $y = a$ of the square there should be, by continuity, a $x \in [0,1]$ such that $f(x,a) \in C$

In a sense, these points should divide the square in two parts, for if there was a "gap" somewhere in the set of such points we could jump from $L$ to $R$ and break continuity.

(EDIT: The question was not correctly formalized, see at the bottom to see the question as it was before)

The question now is whether there is a continuous path making this separation? Formally, is there a continuous path $p : [0,1] \to [0,1] \times [0,1]$ such that $p(0) \in [0,1]\times \{0\}$, $p(1) \in [0,1]\times \{1\}$ (from top to bottom), and that for all $x$, $f(p(x)) \in C$

I'm afraid this might be false for the frontier might be a topologically troublesome space like the topologist's sine curve?

If it is not true in general, is it true in the particular case of $L$ and $R$ being smooth enough, and $f$ being a homotopy? What additional hypothesis would be required to make it true that an homotopy can be split into two "sub"-homotopies?

EDIT: Originally, the question was stated as finding a continuous path $p : [0,1] \to [0,1]$ such that for all $y$, $f(p(y),y) \in C$. This is not possible as proven by Andreas Blass in an Answer: https://mathoverflow.net/q/456226

Answer 1

As far as I can see, there need not be a continuous function $p$ tracing (part of) $C$ as in the question. Consider the piecewise linear path obtained by joining the following four points, by three line segments, in the order listed: $(\frac12,0),\,(\frac14,\frac34),\,(\frac34,\frac14),\,(\frac12,1)$. One can arrange $f$ so that this path is the pre-image of $C$. But one cannot trace this path as the (reflection of the) graph of a continuous $p$. The point is that, when $y$ is near $0$, the point $(p(y),y)$ has to be on the first of my three line segments, when $y$ is near $1$, the point has to be on the third segment, but there is no way to get from the first to the third continuously without following the second line segment, which would make $p$ multivalued.

Answer 2

This addresses the revised question.

There are examples where this is impossible. For example, define subsets of $[0,1]^2$ by $$ \begin{align*} A &= \{(x,y) \mid y > 0, x \leq \tfrac{1}{2} + \tfrac{\sin(1/y)}{4}\}\\ B &= \{(x,y) \mid y > 0, x \geq \tfrac{1}{2} + \tfrac{\sin(1/y)}{4}\}\\ \end{align*} $$ Then $\bar A$ and $\bar B$ are closed subsets covering $[0,1]^2$, with $\{0\} \times [0,1] \subset \bar A$ and $\{1\} \times [0,1] \subset \bar B$, whose intersection $K = \bar A \cap \bar B$ is homeomorphic to a topologist's sine curve. Then "distance from the closed set $K$" is a continuous function, and we can define a function $f: [0,1]^2 \to [0,2]$ by $$ f(p) = \begin{cases} 1 - d(p,K) &\text{if }(x,y) \in \bar A,\\ 1 + d(p,K) &\text{if }(x,y) \in \bar B. \end{cases} $$ Then $f$ is a continuous function $[0,1]^2 \to [0,2] = G$ such that $f(0,-) \in [0,1] = L$ and $f(1,-) \in [1,2] = R$, but the inverse image of $\{1\} = L \cap R$ is the topologist's sine curve and there is no continuous path as desired.

For a positive solution, you probably want to put more assumptions on $f$. For example, if $f$ is smooth and transverse to $C$, then $f^{-1}(C)$ is a smooth 1-dimensional submanifold of $[0,1]^2$ and you can create such a path by, for example, putting a smooth nonvanishing vector field on $f^{-1}(C)$ so that the associated flow produces a path. (This is not completely trivial. To do this, you have to argue that there exists some point in $[0,1] \times \{0\}$ such that the associated flow line end on $[0,1] \times \{1\}$; this is not unrelated to the Jordan curve theorem.)