Let $U$ be an open set in $\mathbb{C}$. Let $f_0, f_1:[a,b] \rightarrow U$ be (path) homotopic piecewise continuously differentiable functions with a homotopy $H$ and the additional property $f_0(a) = f_1(a)$, $f_0(b) = f_1(b)$. Is it possible to find a piecewise continuously differentiable homotopy $\tilde{H}:[a,b] \times [0,1] \rightarrow U$ between $f_0$ and $f_1$ such that $$(*) \quad \tilde{H}(a, t) = f_0(a) = f_1(a),\quad \tilde{H}(b, t) = f_0(b) = f_1(b)$$ for all $t \in [0,1]$?

We know that finding a homotopy without ensuring the property $(*)$ is possible, since we can use the compactness of $H([a,b] \times [0,1])$ to find an $\epsilon > 0$ such that the $\epsilon$-neighbour of $H([a,b] \times [0,1])$ is contained in $U$. And from there we can just approximate $H$ using another continuously differentiable function $H'$ to the point that $$\sup_{s,t}||H(s,t) - H'(s,t)|| < \epsilon.$$ Lastly, we glue the homotopy $H'$ with some linear homotopes from $f_0(t)$ to $H(t, 0)$ and $f_1(t)$ to $H(t,1)$. Hence we get a continuously differentiable homotopy but still likely without the property $(*)$.

How do we find $\tilde{H}$? Is it possible to modify $H$ to get $\tilde{H}$? Or do we need another approach?

  • $\begingroup$ What assumptions are given regarding the domain and range of $H$ in your post? In the context of your first two sentences it seemed clear that $H : [a,b] \times [0,1] \to U$, but then the later paragraph seems not to assume that. However, if you simply meant $H : [a,b] \times [0,1] \to \mathbb C$ then I would say that $H$ plays little role in your problem, because all paths in $\mathbb C$ are homotopic in $\mathbb C$. So it's hard to figure out what you're really asking... $\endgroup$
    – Lee Mosher
    Apr 29, 2023 at 15:15
  • $\begingroup$ @LeeMosher I was talking about $U \subset \mathbb{C}$. Not all paths are homotopic in $\mathbb{C}$ if you also give conditions on the endpoints in order to call two paths homotopic $\endgroup$
    – hteica
    Apr 29, 2023 at 16:42
  • $\begingroup$ In that case you should hit the edit button to clear up two things in your post: to make the domain and range of $H$ explicit; and to emphasis the path homotopy concept which you did not mention in your post. $\endgroup$
    – Lee Mosher
    Apr 29, 2023 at 17:57
  • $\begingroup$ @LeeMosher the domain was always explicit, and in the post its said that $H$ is a homotopy with an additional restrictions on the endpoints $(*)$ which is the definition of path homotopy. I'm sorry I didn't mention the term path homotopy, I simply never used it, but will start to from now on. $\endgroup$
    – hteica
    Apr 29, 2023 at 18:18
  • $\begingroup$ Well, I'll try once more: is the homotopy $H$ given as $H : [a,b] \times [0,1] \to \mathbb C$? Or is it instead given as $H : [a,b] \times [0,1] \to U$? $\endgroup$
    – Lee Mosher
    Apr 29, 2023 at 18:37

1 Answer 1

  1. If you first arrange to satisfy condition (*) in the category of continuous homotopies in $\Omega$, you get a continuous map on $D=[0,1]\times S^1$ into $ \Omega$ which by uniform continuity can be sampled on a triangular grid on $D$ and interpolated by a piece-wise linear map that still takes values in $\Omega$. Note that this map will satisfy the condition of being piece-wise differentiable. If you want the map to be even smoother than that and explictly constructed, you can look at splines (e.g. piecewise-polynomial maps.)

  2. In the category of continuous homotopies, see this discussion for details.loop homotopy


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