The trefoil does not bound a smooth proper disk in the four-ball

Apply van Kampen to the diagram complement, split above and below the projection plane and use small crossing balls. The upper region has one meridian generator for each diagram arc; attaching each crossing ball identifies the outgoing under-meridian with the conjugate of the incoming under-meridian by the over-meridian, since sliding its based normal circle past the over-strand traverses that over-meridian and its inverse. Label the three arcs $a,b,c$ cyclically so that the relations are $c=ab{a}^{-1}$, $a=bc{b}^{-1}$, $b=ca{c}^{-1}$. These are the three positive crossings of the standard trefoil diagram. Substitute the first relation into the second to get $aba=bab$; the third then follows from these two. The knot exterior thus has group $G=⟨a,b\mid aba=bab⟩$, with $a,b$ meridians and the abelianization taking each to $1\in \mathbb{Z}$. This supplies the particular diagram computation rather than importing a general Wirtinger theorem.

The unbranched double cover of the exterior is the kernel $K$ of the mod-two meridian homomorphism $G\to \mathbb{Z}\mathrm{/}2$, by the based-path construction of the cover. To extend it across the branch knot, glue the cover of its solid-torus neighbourhood by squaring each normal disk coordinate. The meridian upstairs then projects to the square of a meridian downstairs. Van Kampen kills its normal closure in $K$. Killing all such lifted meridians gives exactly the kernel of the parity map on $G\mathrm{/}⟨\text{\hspace{0.17em}⁣}⟨a^2,b^2⟩\text{\hspace{0.17em}⁣}⟩$: the normal subgroup is generated by conjugates of meridian squares, and all those conjugates lie in $K$; both choices of lift are included when the boundary covering torus is filled. Thus the branched boundary cover has group equal to that kernel.

The quotient has presentation $⟨a,b\mid a^2=b^2=1,aba=bab⟩$. With involutions the last relation is $(ab{)}^3=1$. Reducing words by $a^2=b^2=1$ and $(ab{)}^3=1$ leaves at most six possibilities, namely $1,ab,(ab{)}^2,a,(ab)a,(ab{)}^{2}a$. Sending $a,b$ to adjacent transpositions in the permutation group of three letters satisfies the relations and yields all six permutations, so the quotient has exactly six elements. Parity is their permutation sign; its kernel consists of the three powers of $ab$ and is cyclic of order three. Consequently the branched double cover $M$ has $H_1(M;\mathbb{Z})=\mathbb{Z}\mathrm{/}3$, by abelianization of its fundamental group.

If the trefoil bounded a smooth proper disk, the disk-cover lemma would give a compact oriented rational homology four-ball $W$ with boundary $M$. The square-order lemma would force $\mathrm{\mid }{H}_1(M)\mathrm{\mid }$ to be a square. Its value is $3$, contradiction.