I am working through a proof in differential forms that if the curves $c_0$ and $c_1$ are homotopic and the homotopy $F:[0,1]^2\to U\subset\mathbb{R}^n$ is $C^1$ and $\omega^1$ is closed $1-$form on $U$, then the line integrals along $c_0$ and $c_1$ coincide.
In the proof it is stated but not explained why $F|_{(0,s)}^* \omega^1 = 0$ after it states that $F(0,s) = c_0(0) = c_1(0)$ is independent of $s$. The only thing I can think is that $F(0,s)$ is constant and so its derivative is $0$ for all $s\in[0,1]$. But I can't extend to why this will make any 1-form uniformly equal to $0$.
The general statement is that if $I \subset \mathbb R$ is an open interval, and if $\gamma : I \to \mathbb R^n$ is a constant path, then for any $1$-form $\omega$ on $\mathbb R^n$ we have $\gamma^* \omega = 0$. To see why, pick $s \in I$, and pick a tangent vector $v \in T_s I \approx \mathbb R$ and let's evaluate: $$\gamma^* \omega(v)= \omega(D\gamma|_s(v)) = \omega(\vec 0) = 0 $$ The fact used here is that since $\gamma$ is constant, the linear transformation $D\gamma|_s : \mathbb R \to \mathbb R^n$ is zero, because the velocity vector $\gamma'(s)$ is the zero vector and so $D_\gamma|_s(v) = v \cdot \gamma'(s) = v \cdot \vec 0 = \vec 0$.