Let $(M,d)$ and $(N,\rho)$ be metric spaces. A function $f: M \to N$ is *Lipschitz* if there exists some constant $\kappa \geq 0$ so that $\rho(f(x),f(y))$ is smaller than $\kappa d(x,y)$ for all points $x,y \in M$.


If $f$ and $g$ are two Lipschitz maps from $M$ to $N$, a *Lipschitz homotopy* between them is a Lipschitz map $H: M \times [0,1] \to N$ for which $H(x,0) = f(x)$ and $H(x,1) = g(x)$ for each $x \in M$. Here we assume that the metric $\Delta$ on $M \times [0,1]$ is given by


$$\Delta\left((x,t),(y,s)\right) = \max\left\{d(x,y),|s-t|\right\}.$$


Here's the question which has been driving me nuts:



> 
> Is there an example of metric spaces $M, N$ and Lipschitz functions $f,g:M \to N$ which are homotopic, but not Lipschitz-homotopic?
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> 


If there is any justice in the world, the answer is an easy "yes". And in this case, I ask



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> Does the answer continue to be "yes" if $M$ and $N$ are assumed compact?
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When does a pair of homotopic Lipschitz functions fail to admit a Lipschitz homotopy?
There is no "justice" in the world of arbitrary metric spaces:-)
Let your first space $M$ be the ordinary circle with its ordinary metric, and the second
space $N$ be homeomorphic to a cylinder with two border circles. Let the metric
be usual on the border circles, but very rough (bumpy) on the middle part of the cylinder.
You can certainly make it so rough that intermediate curves between the border
circles will simply have infinite length...
And of course you can make this example compact. 


