Prove some intermediate curve hits the point Consider two simple smooth curves in $\mathbb{R}^2$ parametrised by $f:[0,1]\rightarrow \mathbb{R}^2$ and $g:[0,1]\rightarrow \mathbb{R}^2$ and with $f(0)=g(0)$ and $f(1)=g(1)$. How can we prove that a point $x\in\mathbb{R}^2$ "inside the curves" (the black dot below) is hit by some intermediate curve $H(s,\cdot) = sg+(1-s)f$?

My own efforts: I think it is hard to formalise the question: one could characterise $x$ by its winding number with respect to the curve $f\circ(-g)$ as this would be 1 (or -1 depending on the orientation). Let us denote this $wnd(x,f\circ(-g))$. Then $F(s)=\int_{\mathbb{R}^2} wnd(x,H(s,\cdot)\circ(-g))dx$ is a continuous function wrt. $s$. Furthermore, it is $F(1)=0$ and equal to the area between $f$ and $g$ for $F(0)=0$. So how could  $wnd(x,H(s,\cdot)\circ(-g))$ change for some $s=s_0$ without there being a $t_0$ such that $H(s_0,t_0)=x$.
Another possibility is to use the "crossing rule" somehow see this post: the fact that two points on adjacent connected components of the plane (without the curve) differ in winding numbers by exactly 1. However, here we are dealing with a moving curve not a moving point.
 A: Let's consider the slightly more general scenario that $f,g\colon[0,1]\rightarrow\mathbb{R}^2$ are continuous curves, such that $f(0)=g(0)$ and $f(1)=g(1)$. Let $x\in\mathbb{R}^2$ be, such that $w(x,g^{-1}f)\neq0$ ($w$ is the winding number and my $g^{-1}f$ means the same thing as your $f\circ(-g)$ , I just prefer this notation for curves). The issue with what you write is that $w(x,g^{-1}H(s,-))$ is only defined for $x\in\mathbb{R}^2$, which don't lie on the image of the curve $g^{-1}H(s,-)$, so your $F$ is not the integral of a continuous function. This observation actually allows us to reverse the situation, though. Namely, we want to show there is an $s\in[0,1]$, such that $x$ lies on the image of $g^{-1}H(s,-)$. So, if we assume this not to be the case, then $s\mapsto w(x,g^{-1}H(s,-))$ is a well-defined continuous function on $[0,1]$, which takes integer values, is non-zero at $s=0$ by assumption, but zero at $s=1$, since $g^{-1}g$ is a contractible curve, contradicting connectedness of $[0,1]$. Thus, there is an $s\in[0,1]$, such that $x$ lies on the image of $g^{-1}H(s,-)$, but it doesn't lie on the image of $g^{-1}$ by assumption, so it has to lie on the image of $H(s,-)$, as desired.
In the language of algebraic topology, the assumption that $g^{-1}f$ has non-zero winding number around at $x$ means that $g^{-1}f$ is not contractible in $\mathbb{R}^2\setminus\{x\}$. Now, $H$ is a homotopy in $\mathbb{R}^2$ between $g^{-1}f$ and $g^{-1}g$, both curves lie in $\mathbb{R}^2\setminus\{x\}$, the former is non-contractible and the latter is contractible in $\mathbb{R}^2\setminus\{x\}$, so the homotopy $H$ cannot be a homotopy in $\mathbb{R}^2\setminus\{x\}$, whence there has to be $(s,t)\in [0,1]\times[0,1]$, such that $H(s,t)=x$.
A: Draw a simple closed curve $C$ from $f(0)$ to $f(1)$ which passes through $x$. As $C$ is traversed from $f(0)$, define the value of a point on $C$ to be $0$ if it is reached before $x$ and $1$ otherwise.
For any value of $t$, the straight line from $f(t)$ to $g(t)$ must cross $C$ at one or more points. Define these points to form the set $S(t)$.
$S(0)$ consists only of point(s) with value $0$ whereas $S(1)$ consists only of point(s) with value $1$. By continuity, this is only possible if some $S(t)$ contains $x$.
