5
$\begingroup$

According to the definition of homotopy between two maps $f,g : X \to Y$ we need a continuous map $F : X \times [0,1] \to Y$ such that $F(x,0) = f(x)$ and $F(x,1) = g(x)$. Most examples I've seen in the literature however tend to skip over the continuity part of $F$ in examples.

So for example take $X$ a convex subset of $\mathbb{R}^n$, fix $x_0 \in X$ and define maps $X \to X$ by $f(x) = x_0$, i.e. $f$ is the constant function and the identity map $\mathrm{id}$. Then for a homotopy between $f$ and $\mathrm{id}$ we clearly want $F(x,t) = tx_0 + (1-t)x$ (obviously could generalise this to any $f$ and $g$). It seems non-trivial to me to prove that $F$ is continuous (the rest of it is straight forward).

I can formally solve to get an $F^{-1}$ for this but I'm not sure how that might help. I get $$ F^{-1}(y) = \left\{ \begin{array}{rl} \{ (x, 1) : x \in X \} \cup \{ (x_0, t) : t \in [0,1) \} & \text{if } y = x_0 \\ \left\{ \left( \frac{y-tx_0}{1-t}, t \right) : t \in [0,1) \right\} & \text{otherwise.} \end{array} \right. $$ but it seems tricky to then apply this to say the open ball about $x_0$ in $X$ say.

I also thought maybe trying to break it down into simpler maps but I can't see this either at the moment, for example I can't break it into maps $X \to X$ and $[0,1] \to X$ as then their product would be a map $X \times [0,1] \to X \times X$ which is of course not what I want.

Any help appreciated. I have a feeling it's not difficult I just can't see it right now.

Steve

$\endgroup$

2 Answers 2

4
$\begingroup$

Working directly from the definition of continuity to check that functions are continuous is going to be fiddly. But you can use known results about continuity to break down the problem. For example:

  • If $f, g : X \to \mathbb{R}^n$ are continuous then the function $x \mapsto f(x) + g(x)$ is continuous;
  • If $f : X \to \mathbb{R}$ and $g : X \to \mathbb{R}^n$ are continuous then the function $x \mapsto f(x)g(x)$ is continuous.

So here the problem is reduced to proving that constant functions and the projection functions are continuous, since the function $X \times [0,1] \to \mathbb{R}^n$ given by $(x,t) \mapsto tx_0 + (1-t)x$ can be written as sums and products of functions of these types.

But checking that constant functions and projection functions are continuous is easy peasy.

$\endgroup$
1
  • $\begingroup$ Of course, many thanks. $\endgroup$
    – Strottos
    Dec 23, 2013 at 23:03
3
$\begingroup$

If $X$ is a subset of $\mathbb R^n$, you can use the fact that $\mathbb R^n$ is a topological group (actually a Lie group), so that the sum and product (product in $\mathbb R^n ; n>2$ means scalar product $\mathbb R \times \mathbb R^n \rightarrow \mathbb R^n$) are both continuous operations.

NOTE: An important temptation to avoid is that of trying to show the continuity of $f(x,y)$ by showing the continuity of both $f(x,.)$ and $f(.,y)$ separately, since this does not imply the continuity of $f(x,y)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .