Continuous, injective map between annuli, but with a "hole" in the image In $\mathbb R^n$ let $B_r$ be the open ball with center zero and radius $r$. For $r\in (0,1)$ let $A_r = \overline{B_1}\setminus B_r$. Let $r,s\in (0,1)$ and assume that $F : A_r\to A_s$ is continuous and injective such that $F(\partial B_1) = \partial B_1$ and $F(\partial B_r) = \partial B_s$. Is it possible that the image of $F$ contains a hole in $A_s$, i.e., $F(A_r) = A_s\setminus U$, where $U$ is a connected open set?

No, that's not possible. Let's pick $r = s$, and in fact work with an annulus $A$ of inner radius $1$ and outer radius $2$. And let's pick a point $P$ in the set $U$, so that $P$ is a point of $A$ such that $P \notin F(A)$. So we have our map,
$$
F : A \to A
$$
whose image misses $P \in A \subset \Bbb R^2$. Define
$$
\gamma_c
$$
to be a path starting at $(1,0)$ and travelling in a straight line to $(1+c, 0)$, then traversing a circle of radius $1+c$ counterclockise, and then returning to $(1,0)$, so that
$$
\gamma_c(t) = \begin{cases}
(1 + 3ct, 0) & 0 \le t \le \frac13 \\
((1+c)\cos(6\pi(t-\frac13)), (1+c)\cos(6\pi(t-\frac13))) & \frac13 \le t \le \frac23\\
(1 + t - 3(t-\frac23), 0) & \frac23 \le t \le 1
\end{cases}
$$
Then $\gamma_0$ and $\gamma_1$ are homotopic loops in $\pi_1(A, a)$, where $a = (1, 0)$. And $\gamma_c$ is homotopic to these for all values of $0 \le c \le 1$. In particular, the loop $\alpha$ defined by $\gamma_0$ followed by $\gamma_1$ is null-homotopic in $\pi_1(A,a)$. That means that $F \circ \alpha$ is nullhomotopic in $F(A)$, hence (by inclusion) in $\pi_1(\Bbb R^2 \setminus \{P\}, F(a)) = \Bbb Z$.
But $F \circ \alpha$ winds once around the point $P$ (OK, that takes a little proving, but not much), hence represents a nonzero element of $\pi_1(\Bbb R^2 \setminus \{P\}, F(a))$, which is impossible, because $F_\star$ is a homomorphism of groups, and cannot send $0$ to a generator.
