Is there a simple proof of the fact that: 



> 
> If $A\subset S^3$ is homeomorphic to $S^1$, then there is a circle $B$
>  embedded into $S^3\setminus A$ that such that the circles $A$ and $B$
>  are linked with the linking number $1$?
> 
> 
> 


If $A$ is smoothly embedded, than it is easy. The problem is that the general topological embedding can be very complicated and I do not see a simple geometric explanation of this fact. I know that $H\_1(S^3\setminus A)=\mathbb Z$ (Corollary 1.29 in Vick's *Homology Theory*) so the generator of the homology group should be a circle. 


**Edit 1:**
This is actually true as John Klein pointed out in his answer below, but I do not know the answer to the higher dimensional version of the problem stated below. Only a partial answer is given in comments below.


The question can be then generalized to higher dimensional spheres. 



> 
> If $A\subset S^n$ is homeomorphic to $S^k$, there should be a
>  topological sphere in $S^n\setminus A$ of dimension $n-k-1$ that links
>  $A$ with the linking number $1$.
> 
> 
> 


**Edit 2.** A counterexample is given below in my answer.



Linking topological spheres
Let $C = S^3 \setminus A$. Alexander duality says that 
$$
H\_1(C) \cong H^1(A) \cong \Bbb Z\, .
$$
Let $\alpha: S^1 \to C$ be any map representing a generator of $H\_1(C)$ (every first homology class is spherical by the Hurewicz theorem). By homotopical approximation we can assume $\alpha$ is a smooth embedding. We can also assume without loss in generality that the image of $\alpha$ misses the north pole of $S^3$. Identify the complement of the north pole with $\Bbb R^3$. Set $B= \alpha(S^1)$. Then $A\amalg B\subset \Bbb R^3$. The degree of the map 
$\require{AMScd}$
\begin{CD}
\ell: A \times B @>>> S^2
\end{CD}
given by $(x,y) \mapsto (x - y)/|x - \alpha(y)|$ is the linking number of $A$ with $B$ by definition. On the other hand, this map has degree one (after choosing appropriate homology generators). 


The point is that the pushforward of a generator 
\begin{CD}
H\_1(S^1) @>\alpha\_\ast >> H\_1(C)
\end{CD}
coincides with the degree of $\ell$. 


How can we check this? Well, assuming $A$ has a nice regular neighborhood, we could redefine $C$ as the complement of that neighborhood. Then $\ell$ can be redefined as the degree of the map
$$
A \times C \to S^2
$$
again given by the same formula,
where we are assuming our new $C$ misses the north pole of $S^3$.
Alexander duality says that the induced slant product pairing 
$$
H\_1(A) \otimes H\_1(C) \to H\_2(S^2) = \Bbb Z
$$
is non-singular, so the degree is $\pm 1$.


Even if $A$ fails to have a nice regular neighborhood, we can assume it misses the north pole $x$ and that $C:= S^n \setminus A$ misses another point $y$ of $S^n$. Identify $S^n \setminus x \cong \Bbb R^n \cong S^n \setminus y$. Then, similarly, we obtain a map
 $$
A\times C \to S^2
$$
and there a linking map $A\times B\to S^2$. With these changes, the argument proceeds as before.


