Deformation retraction restricted to a path Let $X$ be a topological space with subspace $Y$. Suppose that we have a (strong) deformation retraction onto $Y$. That is, a continuous map $F:X\times [0,1]\to X$ such that for all $x\in X$, $F(x,0)=x$, $F(x,1)\in Y$, and for all $y\in Y$ and all $t\in [0,1]$, $F(y,t)=y$.
Is it true that for any $x\in X$ and $t\in [0,1]$, $F(F(x,t),1)=F(x,1)$? It corresponds to my intuition regarding deformation retracts as paths $F(x,\cdot):I\to X$ in $X$ such that all points on the path have the same endpoint in $Y$. However I was unable to prove the result. Any thoughts on this? Any help is much appreciated.

Thank you for your answers. Indeed I now realize that the conjecture is false. In fact, I came up with a quite a silly (but easy) counterexample myself:
Let $X=\{a,b,c\}$ with trivial topology, $Y=\{a,b\}$. Define $F:X\times [0,1]\to X$ (continuous because $X$ has the trivial topology) by
$$
F(a,t)=a,\quad F(b,t)=b \quad \mbox{ for all } t\in[0,1],\quad F(c,t)=\begin{cases}
c &\mbox{ if }\quad t=0\\
a &\mbox{ if }\quad t\in(0,1)\\
b &\mbox{ if }\quad t=1
\end{cases}
$$
Now (similarly to the example by Luiz)
$$
F(F(c,t),1)=F(c,1)\iff t\in \{0,1\}
$$
