Let $\alpha$ and $\beta$ be two homotopic paths in a path connected topological space $X$. Let $\alpha(0)=x_0$ $\alpha(1)=x_1$ $\beta(0)=x_2$ and $\beta(1)=x_3$. Let $p:X\rightarrow Y$ be a continous map to another topological space $Y$ such that $p(x_0)=p(x_2)=y_0$ and $p(x_1)=p(x_3)=y_1$ where $y_0,y_1 \in Y$.
Is $p\circ\alpha$ path homotopic to $p\circ\beta$ ?
If $F:I\times I \rightarrow X$ is a homotopy between $\alpha$ and $\beta$ then $F(s,0)=\alpha(s)$ and $F(s,1)=\beta(s)$ for all $s\in I$. So $p\circ F$ is a homotopy between $p\circ\alpha$ and $p\circ\beta$. However will this be a path homotopy? That is will $p\circ F(0,t)= y_0$ and $p\circ F(1,t) = y_1$ ?
Take $X = \mathbb{R}$, $Y = S^1$ the unit circle and $p(x) = \exp(2 \pi i x)$. As $X$ is contractible, every two paths are homotopic. However, taking $\alpha(t) = 0$ and $\beta(t) = t$ ($t \in [0,1]$), the two paths $p \circ \alpha$ and $p \circ \beta$ will not be path homotopic.