I have a question in Capter 15 of Peskin & Schroeder.
The gauge transformation here in its infinitesimal form: \begin{cases} \psi(x) \to V(x)\psi(x) \quad \quad \quad \quad \quad \quad \,\,\,\, \text{(15.41)} \\ V(x)=1+i\alpha^a(x)t^a+ \mathcal{O}(\alpha^2) \quad \quad \text{(15.42)} \\ A_{\mu}^a \to A_{\mu}^a +\frac{1}{g} \partial_{\mu}\alpha^a +f^{abc}A_{\mu}^b \alpha^c \quad \,\,\,\, \text{(15.46)} \end{cases} When $\tilde{s}$ is a parameter of the path $P$, running from 0 at $x=y$ to s at $x=z$ and $P{}$ denotes path-ordering, the Wilson line is written $$ U_p(z(s),y)=P \biggl\{ \exp \left[ ig\int_0^s d\tilde{s}\frac{dx^{\mu}}{d\tilde{s}}A_{\mu}^a(x(\tilde{s}))\,t^a \right] \biggl\} $$
By an analogy with the propagator (4.23) of the time-ordered exponential, $U_P$ is the solution of a differential equation $$ \begin{align}\frac{d}{ds}U_p(x(s),y) &=\left( ig\frac{dx^{\mu}}{ds}A_{\mu}^a(x(s))\,t^a \right) U_p(x(s),y). \tag{15.57} \\ \Leftrightarrow ~~~~\frac{dx^{\mu}}{ds} D_{\mu} U_p(x,y) &=0 \tag{15.58} \end{align}$$
In the following this book is going to show $$ U_p\left(z,y,A^V\right)=V(z)U_p(z,y,A)V^{\dagger}(y) \tag{15.59} $$ where $A^V$ is the gauge transform of $A$.
And $$ D_{\mu}\left(A^V\right)V(x)=V(x)D_{\mu}(A) \tag{15.60} $$ is proved in its infinitesimal version before.
This relation implies that the right-hand side of (15.59) satisfies (15.58) for the gauge field $A^V$ if $U_P(z,y,A)$ satisfies this equation for the gauge field $A$. But the solution of a first-order differential equation with a fixed boundary condition is unique. Thus, if $U_P(z,y)$ is defined to be the solution of (15.57) or (15.58), it indeed has the transformation law (15.59).
I think that the evolution along the path $x^{\mu}$ (or the parameter $s$)
and the evolution in terms of the gauge transfomation are completely different issues.
Why does this book claim as the last line of this paragraph?