= Solution
A <Wirtinger presentation> from a connected <knot diagram> has one generator per arc and one relator per crossing, with one relator redundant. Its presentation complex is a finite two-dimensional <CW complex> $X$ with one zero-cell, $n$ one-cells, and $n-1$ two-cells, and the usual diagrammatic construction gives a <homotopy equivalence> $X\simeq E_K$.
The <abelianization> of the <knot group> is $H_1(E_K;\mathbb Z)\cong\mathbb Z$, generated by a <meridian of a knot>. Every homomorphism to the <cyclic group> $\mathbb Z/2$ factors through this abelianization, and reduction modulo two is its unique surjection. Thus the requested map $\alpha$ is unique.
Its kernel determines a two-sheeted <covering space> $\widehat X$. The nontrivial <deck transformation> acts on cellular chains and homology, giving them module structures over the <group ring>
$$
\widehat R=\mathbb Z[\mathbb Z/2]\cong\mathbb Z[t]/(t^2-1).
$$
Lift one copy of every cell of $X$; its two deck translates form a free $\widehat R$-basis. Hence $C_0\cong\widehat R$, $C_1\cong\widehat R^n$, and $C_2\cong\widehat R^{n-1}$. If $\widetilde X$ is the <infinite cyclic cover>, its cellular chains are free over $R=\mathbb Z[t^{\pm1}]$, and imposing $t^2=1$ gives
$$
C_*^{\mathrm{cell}}(\widehat X)\cong C_*^{\mathrm{cell}}(\widetilde X)\otimes_R\widehat R.
$$
For an odd prime $p$, the two <idempotent>[idempotents] $e_+=(1+t)/2$ and $e_-=(1-t)/2$ split the <group algebra>
$$
\mathbb F_p[\mathbb Z/2]\cong R_+\oplus R_-,
\qquad R_+=\mathbb F_p[t^{\pm1}]/(t-1),\quad R_-=\mathbb F_p[t^{\pm1}]/(t+1).
$$
The plus summand is the cellular chain complex of $X$ with $\mathbb F_p$ coefficients, while the minus summand is $C_-=C_*^{\mathrm{cell}}(\widetilde X)\otimes_RR_-$. Therefore
$$
H_*(\widehat X;\mathbb F_p)\cong H_*(E_K;\mathbb F_p)\oplus H_*(C_-).
$$
On the minus summand, the boundary $t-1:C_1\to C_0$ becomes multiplication by $-2$, which is invertible in $\mathbb F_p$, so $H_0(C_-)=0$. After the corresponding cancellation, the remaining square boundary matrix is an <Alexander matrix> specialized at $t=-1$. It is singular over $\mathbb F_p$ exactly when
$$
\Delta_K(-1)\equiv0\pmod p,
$$
or equivalently when $p$ divides the <knot determinant> $\det K$. Thus $H_*(C_-)$ is nonzero exactly in that case.
Finally, $\det K=|\Delta_K(-1)|$ is a nonzero odd integer, so the minus complex is acyclic over $\mathbb Q$. Since a knot exterior has the rational homology of a circle,
$$
\boxed{H_i(\widehat X;\mathbb Q)\cong
\begin{cases}
\mathbb Q,&i=0,1,\\
0,&i\geq2.
\end{cases}}
$$
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