Let $I:=[0, 1]$ and suppose $(\gamma_s)_{s\in I}$ and $(\gamma_s^\prime)_{s\in I}$ are families of paths $\gamma_s, \gamma_s^\prime:I\longrightarrow X^Y$ with values on the space of continuous functions $X\longrightarrow Y$. Suppose all those paths have fixed extremes and each $\gamma_s^\prime$ is concatenable with each $\gamma_s$.
I won't say anything about the topology on $X^Y$ for I won't really need it.
Fix $x\in X$ and define the path $\alpha_x: I\longrightarrow Y$ setting
$$\alpha_x(s):=\left\{\begin{array}{lcl} (\gamma_0^\prime*\gamma_{2s})(x) & \textrm{if}& s\in [0, 1/2]\\ (\gamma^\prime_{2s-1}*\gamma_1)(x)& \textrm{if}& s\in [1/2, 1] \end{array}\right.,$$ where $*$ stands for the usual concatenation of paths. How can I show $\alpha_x$ is homotopic to the path $\beta_x:I\longrightarrow Y$, $$\beta_x(s):=(\gamma_s^\prime*\gamma_s)(x)?$$
Thanks.
Remark. You don't have to worry about continuity. Once the deformation is found I guess it will turn out to be continous, but that will be my issue =)
I guess I found it. First, define $$a(s):=\left\{\begin{array}{lcl} 0& \textrm{if}& s\in [0, 1/2]\\ 2s-1 & \textrm{se}& s\in [1/2, 1]\end{array}\right.\quad \textrm{and}\quad b(s):=\left\{\begin{array}{lcl} 2s& \textrm{if}& s\in [0, 1/2]\\ 1 & \textrm{se}& s\in [1, 1/2]\end{array}\right..$$ Then, define $h_x:I\times I\longrightarrow Y$ setting: $$h_x(\varepsilon, s):=(\gamma^\prime_{(1-\varepsilon/2) a(s)+(\varepsilon/2) b(s)}*\gamma_{(1-\varepsilon/2)b(s)+(\varepsilon/2)a(s)})(x).$$ Then: $$h_x(0, s)= (\gamma^\prime_{a(s)}*\gamma_{b(s)})(x)=\left\{\begin{array}{lcl} (\gamma^\prime_0*\gamma_{2s})(x)&\textrm{if}& s\in [0, 1/2]\\ (\gamma^\prime_{2s-1}*\gamma_1)(x)&\textrm{if}& s\in [1/2, 1]\end{array}\right.=a_x(s),$$ whereas $$h_x(1, s)=(\gamma^\prime_{(a(s)+b(s))/2}*\gamma_{(a(s)+b(s))/2})(x)=(\gamma^\prime_s*\gamma_s)(x)=\beta_x(s).$$ Hope it's right. If someone could check I'd be pleased.