I've encountered the following problem that I'm sure someone more topologically inclined can answer:


Say that a homotopy of maps $f:X\times[0,1)\to Y$ between two compact smooth manifolds $X$ and $Y$ is good (I don't know the exact term for it if there is one) if the homotopy is constant in time outside of some coordinate chart in $Y$, in which case we'd say the homotopy is supported in the chart.


Suppose you have two continuous maps $f\_0$ and $f\_1$ between compact smooth manifolds $X$ and $Y$ which are homotopic. Cover $Y$ with finitely many coordinate charts $U\_\alpha$. My question is the following:


Are $f\_0$ and $f\_1$ homotopic with respect to a finite sequence of good homotopies supported in the elements $U\_\alpha$ of the covering? Can the number of good homotopies required be uniformly bounded by topological data like the homotopy class of $f\_0$ or the number of elements in the covering?


I am interested in this question because I have a functional which I can get a good estimate for if the above is true.



Are maps homotopic with respect to a uniform number of local homotopies
Yes, you can always connect $f\_0$ and $f\_1$ by a finite sequence of good homotopies supported in any given open cover (and in fact, this sequence of good homotopies can be chosen to be homotopic to $f$ relative to $X\times \{0,1\}$). By pulling back the $U\_\alpha$ to an open cover of $X\times [0,1]$ via $f$, it suffices to show this in the universal case where $Y=X\times [0,1]$ and $f$ is the identity. By refining the open cover, we may assume we have an open cover consisting of all the sets of the form $V\_i\times (n\epsilon,(n+3)\epsilon)$ for some finite open cover $\{V\_1,\dots,V\_m\}$ of $X$ and some $\epsilon>0$. Let $\{\varphi\_i\}$ be a partition of unity subordinate to $\{V\_i\}$. Then you can connect $f\_0$ to $f\_1$ through good homotopies as follows. First, homotope $f\_0:x\mapsto (x,0)$ to the map $x\mapsto (x,\epsilon \varphi\_1)$ through the homotopy $(x,t)\mapsto (x,t\epsilon\varphi\_1)$. Then homotope to $x\mapsto(x,\epsilon(\varphi\_1+\varphi\_2))$ similarly, and so on; after $m$ steps you will have reached $x\mapsto (x,\epsilon\sum \varphi\_i)=(x,\epsilon)$. Now repeat to get to $x\mapsto (x,\epsilon+2)$, and so on until you reach $x\mapsto (x,1)$.


On the other hand, you can't get any uniform bound on the number of steps needed using only $f\_0$ or the open cover. For instance, consider the case where $X=Y=S^1$, $f\_0$ is a constant map, and $f\_1$ winds around $n$ times in one direction and then $n$ times in the opposite direction. By lifting to the universal cover, it is not hard to show that it takes at least $n$ good homotopies to get from $f\_0$ to $f\_1$ (for any covering of $Y$ by charts).


