Linear isomorphism determined by parallel transport 
Suppose $ \nabla $ is an affine connection on a differentiable manifold $ M $, $ 
p, q\in M $, $ c $ is a smooth curve connecting $ p $ and $ q $, then the parallel transport along $ c $ gives us a linear isomorphism $$ P^{c}:T_{p}M\longrightarrow T_{q}M ,$$ and $ P^c $ is uniquely determined by $ \nabla $.

I read this paragraph from my textbook. What confused me is what exactly $ P^c $ looks like such that $ P^c $ is a linear isomorphism?
Edit: I want the explicit expression for $ P^c $.
 A: Let us take local coordinates at $p$ denoted by $x_1,\ldots,x_n$. Let $v\in T_pM$, and write $v=v^i\partial/\partial x^i$. Let $\Gamma_{ij}^k$ be the Christoffel symbols of $\nabla$ in these coordinates. Let $\gamma=(\gamma_1,\ldots,\gamma_n)$ be a smooth path starting at $p$. A vector field $X$ along $\gamma$ can be written as $$X(t)=X^i(t)\frac{\partial}{\partial x^i}.$$ Such a vector field is parallel with respect to $\nabla$ if it solves the equation 
$$\begin{align}0&=\frac{DX}{dt}\\&=\dot{X}^k\frac{\partial}{\partial x^k}+\dot{\gamma}^iX^j\Gamma_{ij}^k\frac{\partial}{\partial x^k}\\&=\left(\dot{X}^k+\dot{\gamma}^iX^j\Gamma_{ij}^k\right)\frac{\partial}{\partial x^k}\;.\end{align}$$
Moreover, it is the parallel transport of $v$ if it satisfies, in addition, the initial value condition $X^i(0)=v^i$. 
In general, it is practically impossible to write an explicit expression for parallel transport, as it is a solution to an ODE, and the coefficients may be selected arbitrarily. However, as demonstrated above, this is a linear ODE of the first order. Hence, the solution space is linear and defined for infinite time. In other words, parallel transport defines a linear isomorphism between the tangent spaces at the two endpoints of a path.
