Homotopy Exercise

I have been trying to write up explicit homotopy for

Exercise 3(iii): If ${\gamma_0, \gamma_1: [a,b] \rightarrow U}$ are closed curves with the same initial point, show that ${\gamma_0}$ is homotopic to ${\gamma_1}$ as closed curves if and only if ${\gamma_0}$ is homotopic to ${\gamma_2 + \gamma_1 + (-\gamma_2)}$ with fixed endpoints for some closed curve ${\gamma_2}$ with the same initial point as ${\gamma_0}$ or ${\gamma_1}.$

But always ended up really messy, and find it hard to show continuity. May someone provide a rigorous example of one direction? Say ($\Leftarrow )$ direction of problem.

I assume that $+$ means the composition of curves. We will prove that $\gamma_1 \sim \gamma_2 + \gamma_1 + (-\gamma_2)$, it is enough since homotopies can be composed and inverted and thus form an equivalence relation. So let us assume given a map $\gamma: [0, 1] \to U$ such that $$\begin{eqnarray} \gamma(t) &=& \gamma_2(3t), &~~~~ & t \in [0;~ 1/3] \\ \gamma(t) &=& \gamma_1(3t-1), & & t \in [1/3;~ 2/3] \\ \gamma(t) &=& \gamma_2(3(1-t)), & & t \in [2/3;~ 1] \end{eqnarray}$$
We define a homotopy $\Gamma:[0,1]^2\to U$ by the rule $$\begin{eqnarray} \Gamma(t,k) &=& \gamma_2(3t+1-k), &~ & t \in [0;~ k/3] \\ \Gamma(t,k) &=& \gamma_1\left(\frac{3t-k}{3-2k}\right), & & t \in [k/3;~ 1-k/3] \\ \Gamma(t,k) &=& \gamma_2(3(1-t)+1-k), & & t \in [1-k/3;~ 1] \end{eqnarray}$$
Obviously it is continuous. When $k=0$ it is equal to $\gamma_1$, when $k=1$ it is equal to $\gamma$.
Why do we use these formulas? It is easy to see if we don't insist on writing everything explicitly. Obviously reparametrization preserves homotopy equivalence, thus $\gamma$ is equivalent to $\gamma_1 + ((-\gamma_2) + \gamma_2)$ by rotation of circle. Now for any curve $s$ it is true that $s + (-s) \sim 0$ where $-s$ is $s$ with reversed parametrization. Geometrically $s + (-s)$ consists of going along a curve on the first half of the path and going in reverse on the second half. Of course we can contract such loop into its base point by paths that reverse earlier. Now it's just linear algebra to write down these homotopies in the original parametrization.