Is the space of gradient-like vector fields contractible? Let $M$ be a compact manifold (without boundary) and let $f:M\to \mathbb{R}$ be a fixed Morse-function.

Question: Is the space $GVect(M,f)$ of all gradient-like vector-fields for $f$ contractible or even convex?

To eliminate any misunderstanding, this is the definition of gradient-like vector fields that I'm using:

Definition: A vector field $X\in Vect(M)$ is called gradient-like for $f$ if
   1.  $\forall q\in M\setminus Crit(f):\; df(q)X(q)<0$.
   2. For each critical point $p\in Crit(f)$ of $f$, $\exists$ Morse coordinate chart $\phi:U\to\mathbb{R^n}$ (i.e. $\phi(p)=0$ and $f\circ\phi^{-1}(x)=f(p)-\sum_{i=1}^k x_i^2+\sum_{i=k+1}^n x_i^2$ for $x\in\phi(U)$, $k=index(p)$) such that $$\phi_*X(x)=d\phi(\phi^{-1}(x))X(\phi^{-1}(x))=(2x_1,\ldots,2x_k,-2x_{k+1},\ldots,-2x_n)^T$$  for $x\in\phi(U)$ (i.e. in this Morse-chart $X$ coincides with the negative gradient of $f$ w.r.t. the euclidean metric on  $\mathbb{R^n}$).

Thank you in advance for any insights.
 A: $GVect(M,f)$ is a convex set in the space of all vector fields. Thus it is contractible.
Edit, since you asked for details: I assume that the Morse charts $\Phi$ are the same for all $X$. 

Let $X,Y$ be gradient-like for $f$, then:

1. $df(q)(t.X(q)+(1-t).Y(q)) = t.df(q)(X(q)) + (1-t).df(q)(Y(q)) < 0$ for $0\le t\le 1$.

2. In the Morse charts the two vector fields look the same.
Second edit: Okay, you want the Morse chart to depend on the vector field.
Maybe, your definition of a gradient-like vector field is equivalent to:
a) At a critical point $p$ we have $X(p)=0$. 

b) Near each critical point $p$, the function $-X(f)= -df(X)$ is Morse with critical point $p$ a minimum. Then you can use $-df(X)$ as the square distance from $p$ for a Riemannian metric, and the Morse chart construction with respect to that Riemannian metric leads to a Morse chart as you require.

c) Off the critical points, we have $df(X)<0$ (i.e. X is transversal to the level sets and points towards lower level sets).      
