Removing the point at infinity preserves the knot group

Since $K$ is compact, choose $R>0$ with $K$ and $b$ in the open ball $D=B_R(0)$. Put $U=S^3\setminus (K\cup \{\mathrm{\infty }\})={\mathbb{R}}^3\setminus K$ and $V=S^3\setminus \bar{D}$. Then $V$ is an open ball about infinity, disjoint from $K$, and $U\cap V=V\setminus \{\mathrm{\infty }\}$. First verify that $U$ is path-connected without assuming a diagram code. Parameterize the smooth embedded circle by $k:S^1\to K$. For $p\in U$, the cone $C_p=\{p+t(k(s)-p):s\in S^1,t\ge 1\}$ has three-dimensional outer volume zero: on each of finitely many smooth coordinate intervals of $S^1$ and each bounded interval $1\le t\le n$, the defining map is Lipschitz on a compact rectangle; subdividing into $O(N^2)$ squares of side $1\mathrm{/}N$ covers its image by balls of radius $O(1\mathrm{/}N)$, whose total three-dimensional volume is $O(1\mathrm{/}N)$. Thus each truncated image has outer volume zero. For each $n$, the finite chart and derivative bounds give an explicit cover volume $A_n\mathrm{/}N$ after mesh size $1\mathrm{/}N$; choosing $N_n>2^{n+2}{A}_n\mathrm{/}ϵ$ makes the $n$th truncated images of both cones have total cover volume below $ϵ2^{-n-1}$. These formula-defined mesh sizes give a cover of $C_p\cup C_q$ with total volume below $ϵ$ for every $ϵ>0$, so the union has outer volume zero without a choice of covers. An open ball contains a cube of positive volume, so choose $r$ in an open ball outside $C_p\cup C_q$. If the segment $[p,r]$ met $K$ at $k(s)$, then $r=p+t(k(s)-p)$ for some $t\ge 1$, contrary to $r\mathrm{\notin }{C}_p$; similarly $[q,r]$ misses $K$. The broken line from $p$ to $r$ to $q$ lies in $U$. Under inversion $J(x)=x\mathrm{/}\mathrm{\parallel }x{\mathrm{\parallel }}^2$, extended by $J(\mathrm{\infty })=0$, the set $V$ is the open ball of radius $1\mathrm{/}R$ about $0$. The radial homotopy $y\mapsto (1-t)y$ contracts it to $0$, so it is nonempty and path-connected. Since this ball is convex, every based loop contracts to its own basepoint by the straight-line homotopy inside the ball; [F3] therefore makes $V$ simply connected. The same inversion identifies $U\cap V$ with $0<\mathrm{\parallel }y\mathrm{\parallel }<1\mathrm{/}R$. Fix $\rho =1\mathrm{/}(2R)$; the puncture-preserving radial homotopy $y⟼((1-t)+t\frac{\rho }{\mathrm{\parallel }y\mathrm{\parallel }})y$ is a strong deformation retraction of this punctured ball onto the sphere $S_{\rho }^2$. The punctured ball is path-connected. Choose $y_0\in S_{\rho }^2$, let $i:S_{\rho }^2\hookrightarrow U\cap V$ be inclusion, and let $r$ be the radial retraction. Since the radial deformation fixes $S_{\rho }^2$ pointwise, [F7] and functoriality make $i_{\ast }$ and $r_{\ast }$ inverse isomorphisms at $y_0$. By [F2], ${\pi }_1(S_{\rho }^2,y_0)$ is trivial. For any other basepoint $y$ in the punctured ball, choose a path $\gamma$ from $y$ to $y_0$ and use $[\alpha ]\mapsto [\bar{\gamma }\ast \alpha \ast \gamma ]$. Concatenation and reparameterization preserve endpoint-fixed homotopies, so this is a well-defined homomorphism. The inverse uses $\bar{\gamma }$; the inserted pairs $\gamma \ast \bar{\gamma }$ and $\bar{\gamma }\ast \gamma$ contract by the piecewise retracing homotopy, so the homomorphism is an isomorphism. Thus every basepoint group of the overlap is trivial; its nonemptiness and path-connectedness with [F3] make $U\cap V$ simply connected.

Choose $b_0\in U\cap V$. The open sets $U,V$ cover $S^3\setminus K$; [F4] applies by the path-connectedness in step 1.1. Since $V$ and $U\cap V$ are simply connected by step 1.1, the special case [F5] shows that inclusion induces an isomorphism ${\pi }_1(U,b_0)\to {\pi }_1(S^3\setminus K,b_0)$.

The inclusion induces the based map by [F6]. Choose a path $\gamma$ in $U$ from $b$ to $b_0$, which exists by step 1.1. The formula $[\alpha ]\mapsto [\bar{\gamma }\ast \alpha \ast \gamma ]$ defines isomorphisms between the fundamental groups at these basepoints by the concatenation and retracing argument in step 1.1. Using the same path in $S^3\setminus K$ makes the square with inclusion commute, since postcomposing the concatenated path gives the same loop. Thus the inclusion map at $b$ is conjugate by isomorphisms to the isomorphism at $b_0$.