For a topological space $X$ and a positive integer $k\in \mathbb{N}\_{>0}$ let $F\_k(X):= \{ (x\_1,\ldots,x\_k)\in X^k |x\_i\neq x\_j \text{ for } i\neq j \}$ be its $k$-configuration space. 
Let $f:M\to \mathbb{R}$ be a Morse function on a compact manifold $M$.
The space $GVect(f)$ of all gradient-like vector fields for $f$ is convex (if a suitable definition of gradient-like is used) and hence contractible.



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> **Question:** Is the $k$-configuration space $F\_k(GVect(f))$ contractible?
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My approach so far:


Let $Q\_m\subset GVect(f)$ denote a set of $m\geq 0$ distinct points.


*Claim:* 


1. $GVect(f)$ is a manifold without boundary;
2. $GVect(f)\setminus Q\_m$ has trivial homotopy groups;
3. $F\_k(GVect(f))$ is a $CW$-complex.


Using parts 1 & 2 of the claim, theorem 2.5 and the proof of theorem 2.7 of [this paper](http://www.math.rochester.edu/people/faculty/jonpak/newbraid.pdf), one can show that the homotopy groups of $F\_k(GVect(f))$ are trivial.
Whitehead's theorem and part 3 of the claim now imply that $F\_k(GVect(f))$ is in fact contractible.


However, I'm not sure if the claim is true (or if it's even sensible) and how one may prove it.


Thank you for any contribution.



Contractibility of a configuration space
It seems that $GVect(f)$ is a manifold without boundary. 
Building on my [answer](https://mathoverflow.net/questions/123991) to your last question (did you prove all of it?) let us argue as follows:
A vector field $X$ is in 
$GVect(f)$ if:


* $X(p)=0$ for each critical point $p$ of $f$. This describes a closed linear subspace $G\_1$ of the Frechet space $\mathfrak X(M)$.
* Near each critical point $p$ the function $df(X)$ is Morse with a maximum at $p$. This is a $C^2$-open condition in $G\_1$: The differential must be transversal to the zero section, at $p$.
* Off the critical points we have $df(X)<0$. This would be a $C^1$-open condition in $G\_1$ if it holds on a closed subset of $M$. We can take the closed subset as the complement of the union of small open neighborhoods of the critical points of $f$. But these neighborhoods depend on $X$. So we have to look at all of them and take the union. Since the union of open sets is open, we are done.


So $GVect(f)$ is open in the Frechet space $G\_1$. 


The rest seems to be done by the comments to your question, by Geoffroy.


