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Let $X = \mathbb R^N$. Let $\Phi$ be the set of continuous contractions of $X$ to a single point. That is, $\Gamma \in C([0,1] \times X, X)$ is in $\Phi$ if

  • $\Gamma(0,u)= u $ for all $u \in X$
  • $\Gamma(1,u) = u_0$ for some $u_0 \in X$
  • $\Gamma(t, u)$ is a homeomorphism of $X$ for $t \in [0,1)$, and $\Gamma(t,u)^{-1}$ is continuous on $[0,1) \times X$

Now, take two points $e_1, e_2 \in X$.

My question: given a continuous curve $\gamma \in C([0,1], X)$ with $\gamma(0) = e_1$, $\gamma(1) = e_2$, does there exist a $\Gamma \in \Phi$ such that $$ \Gamma(t, e_1),\ \Gamma(t, e_2) \in \gamma([0,1]) \quad \forall t \in [0,1] $$ That is, does there exist a contraction of $X$ that restricts to the image of the given curve on $\{e_1, e_2\}?$ Apologies if this is a simple question.

eugenefraxby
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1 Answers1

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I will need two facts for my answer:

  1. If $\gamma: [0,1]\to \mathbb R^n$ is a path such that $\gamma(0)=p\ne q=\gamma(1)$ then there is an injective path $\alpha: [0,1]\to \gamma([0,1])$ such that $\alpha(0)=p, \alpha(1)=q$. (See e.g. here.)

  2. If $u: [0,1)\to S^{n-1}$ is a path in the unit sphere, then there exists a path $h: [0,1)\to SO(n), h(t)=A_t$ such that $A_t(\mathbf e_1)=u(t), t\in [0,1)$ and $A_0=I$, the identity matrix. (This comes from the path-lifting property for the standard fibration of $SO(n)$ over $S^{n-1}$ given by the action of $SO(n)$.)

Suppose now that $\gamma: [0,1]\to \mathbb R^n$ is a continuous injective path. Without loss of generality (after composing with an affine transformation), we may assume that $\gamma(0)=\mathbf 0$ and $\gamma(1)=\mathbf e_1$, the first basis vector. Define $c_t:= ||\gamma(1-t)||$. By injectivity of $\gamma$, $c_t>0$ for all $t\in [0,1)$. Define the unit vector $$ u_t:= c_t^{-1} \gamma(1-t). $$ By (2), there exists a path $h(t)=A_t$ of orthogonal matrices, such that $A_t(e_1)=u_t$ for every $t\in [0,1)$. Define linear transformations $$ f_t:= c_tA_t $$ and set $F(x,t)=f_t(x)$. Then $F(x,0)=x$ for every $x\in \mathbb R^n$. Furthermore, $F$ is continuous and extends continuously to a map $F: \mathbb R^n\times [0,1]$ by $F(x,1)=0$.

There is one more case to consider: $\gamma(0)=\gamma(1)=\mathbb 0$. Then we can take $F(x,t)=(1-t)x$.

Moishe Kohan
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