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?
 A: Here is a proof using invariance of domain theorem. Take your annulus $A_r$ and "double it" gluing two copies $A^\pm$ of $A=A_r$ along boundary spheres. The result is a connected closed (compact and with empty boundary) manifold $M$ (it is homeomorphic to $S^{n-1}\times S^1$ in case you are interested). The mapping $F: A\to A$ yields a continuous injective mapping $DF: M\to M$. Since $M$ is compact and $F$ is continuous, $DF(M)$ is also compact. Since $M$ is Hausdorff, $DF(M)$ is closed. By the invariance of domain theorem (this is where you need homology), $DF(M)\subset M$ is open. Since $M$ is connected, $DF(M)=M$. Thus, $F(A)=A$. qed
Edit. Let $(X,A)$ be a topological space $X$ with a closed subset $A$. The double $DX$ along $A$ is the quotient space of the product
$$
X\times \{0, 1\}
$$
(where $\{0, 1\}$ has the discrete topology)
by the equivalence relation $(a, 0)\sim (a, 1)$ for all $a\in A$. (The above product space is a disjoint union of two copies of $X$. The space $DX$ is informally described as obtained from that disjoint union by gluing two copies of $A$. It is an good exercise to understand the example when $X$ is the closed disk and $A$ is its boundary circle. Then $DX$ is homeomorphic to $S^2$.) Since $A$ is closed, $DX$ is Hausdorff, provided that $X$ is.
(Example: If $X$ is a manifold with boundary and $A=\partial  X$, then $DX$ is a manifold without boundary.)
Suppose that $f: (X,A)\to (Y,B)$ is a continuous map of pairs (i.e. $f(A)\subset B$), then one defines its double by
$$
Df: [(x, i)] \mapsto [(f(x), i)], i=0, 1. 
$$
Here bracket denotes the equivalence class as above. The map $Df$ is well-defined since $f(A)\subset B$. The map $Df$ is always continuous. If $f$ is injective, so is $Df$.
A: 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.
