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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 note that the composition $$ A \subset S^3 \subset S^4 $$ has complement containing $\Sigma C \subset S^4$, the suspension of $C$. Then $A \amalg \Sigma C$ misses a point in $S^4$ which we can take to be the north pole. One can show that the degree of the associated map $$ \ell' : A \times \Sigma C \to S^3 $$ coincides with the degree of the map $\ell$. But the degree of the latter map is +1, again by Alexander duality.

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.

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 $$ H_1(B) \to H_1(C) $$ coincides with the degree of $\ell$.

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 note that the composition $$ A \subset S^3 \subset S^4 $$ has complement containing $\Sigma C \subset S^4$, the suspension of $C$. Then $A \amalg \Sigma C$ misses a point in $S^4$ which we can take to be the north pole. One can show that the degree of the associated map $$ \ell' : A \times \Sigma C \to S^3 $$ coincides with the degree of the map $\ell$. But the degree of the latter map is +1, again by Alexander duality.

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 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). This can be verified when $A\subset S^3$ is a smooth submanifold using the definition of the integral degree of a map (as the preimage of a regular value) that appears in Milnor's in Topology From a Differentiable Viewpoint (see the exercise on the linking number at the end of the book).

In the non-smooth case one needs to analyze what the map $\ell$ does on $H_2$. Here is one possible sketch: Note that the composition $$ A \subset S^3 \subset S^4 $$ is unknotted, so $S^4 \setminus A$ is homotopy equivalent to $S^2$. It is elementary to check that the composite $$ \Sigma B \to \Sigma C \subset S^4 \setminus A $$ gives such an equivalence, where $\Sigma$ means unreduced suspension. But it should be apparent that linking number of $A$ with $\Sigma B$ is then $+1$ and this linking number coincides with the linking number of $A$ and $B$.

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 $$ H_1(B) \to H_1(C) $$ coincides with the degree of $\ell$.