Can I deform a solid cylinder to a standard one?

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Let $D^2$ be the closed unit disc in $\mathbb{R}^2$ and denote by $A\subseteq D^2$ the closed disc with radius $1/2$.

Suppose $F:D^2\times [0,1] \to D^2\times [0,1]$ is an embedding (homeomorphism onto its image) such that:

  • $F$ is height preserving, that is $F(D^2\times\{t\}) \subseteq D^2\times\{t\}$ for every $t\in [0,1]$;

  • the image of $F$ does not intersect the lateral surface $\partial D^2 \times [0,1]$;

  • $F(D^2\times \{0\}) = A\times\{0\}$;

  • $F(D^2\times \{1\}) = A\times\{1\}$.

Is it true that there exists a height preserving homeomorphism $H:D^2\times [0,1]\to D^2\times [0,1]$, extending the identity of the boundary, such that the image of the composition $H\circ F$ is $A\times [0,1]$?