Uniformly continuous homotopy equivalence Suppose $M$ and $N$ are complete metric spaces and $f, g: M \to N$ are uniformly continuous maps between them with common modulus of continuity $m$. Further suppose $f$ and $g$ are homotopy equivalent.
Must there be a a homotopy equivalence $\alpha\colon M \times [0, 1] \to N$ between $f$ and $g$ such that $\alpha$ is uniformly continuous?
If so, can we further require that $\alpha$ has a modulus of continuity $m^*$ where $m^*$ only depends on $m$? Can we require $m^* = m$?
Similarly, if $M$ and $N$ are complete metric spaces which are homotopy equivalent, can we find a uniformly continuous homotopy between them? If so can we say anything about the modulus of continuity which is independent of the spaces (e.g. that they are Lipschitz)?
 A: A counterexample is given by $f,g:\mathbb R\to\mathbb R$, $f(x)=x$,
$$
g(x) = \begin{cases} 0 & x<0 \\ x & x\ge 0 \end{cases} .
$$
Then it's not possible to find a $\delta>0$ such that (let's say) $|\alpha(x,s)-\alpha(y,t)|<1$ for all $|(x,s)-(y,t)|<\delta$ because that would in particular mean that $|f(x)-g(x)|<1+1/\delta$, which is obviously not true for small $x$.
A: I think not.
If $M$ is compact, then $\alpha$ must be uniformly continuous; but even then the modulus of continuity can be impossible to preserve; there is some interpretation on what is meant here, depending on the metric you use on the product $M\times [0,1]$, but we can have that there must exist some $t$ such that $\alpha_t:=x\mapsto\alpha(x,t)$ does not have $m$ as modulus of continuity. For example, take $M$ the circle, $N$ an unduloid, and $f,g$ two short geodesics on two different necks of $N$. Then any homotopy must pass through a long curve at the fat part between the necks, which cannot have exactly the same modulus of continuity (here, $m^*$ can be a multiple of $m$).
If $M$ is not compact even the equi-uniform continuity of $\alpha_t$ can be prevented. Take $M=\mathbb{R}$, and for $N$ the plane with a sequence of hills  centered at points $(0,n)$ where $n$ runs over $\mathbb{Z}$ and contained in a band $(\lvert y\rvert \le 1)$. Make the hills very steep and very high when $z$ is large. Let $f = x\mapsto (2,x)$ and $g = x\mapsto (-2,x)$. Then any homotopy from $f$ to $g$ has to incur high distortion to pass through the farther hills.
If $M$ and $N$ are homeomorphic, it might be impossible to get any modulus of continuity for an homeomorphism: a classical operation to see this is snowflaking: take $M=N$ but metric $d$ on $M$ and $d^q$ on $N$, where $q\in(0,1)$. Then they have different dimensions, ruling out a Lipschitz homeomorphism; and you can compose $d$ with an arbitrarily bad concave modulus of continuity to prevent any given modulus to be achieved, so nothing uniform over all metric spaces can be expected.
For homotopy equivalence, you can do this construction on a circle; a map of non-vanishing degree will incur the same distortion as a homeomorphism.
To prevent even uniformly continuous homotopies, one has to look for non-compact spaces again. I guess that using a loch-ness monster obtained by attaching a circle to the line at each integer point could do the trick. For $M$, make the metric give length $1$ to all circle. For $N$, make it give length $\lvert n\rvert$ to the circle at $n$. Any homotopy $M\to N$ must find a way to map a circle of $M$ to each of $N$, and this should prevent uniform continuity.
