A nontrivial Whitney circle in the fundamental group blocks cancellation

Form $X$ by removing a small $6$-ball disjoint from $A=S^3\times \{y_0\}$ and $C_1\cup C_2$ in the product, removing a ball from $S^1\times S^5$, and identifying their boundary $5$-spheres by an orientation-reversing diffeomorphism. This explicitly defines the smooth oriented connected sum. Removing either ball does not change the fundamental group: apply van Kampen to the punctured manifold and the ball, with collar overlap homotopy equivalent to the simply connected $S^5$. The same theorem across the neck gives ${\pi }_1(X)=1\ast \mathbb{Z}=\mathbb{Z}$. Here $S^3,S^5$ are simply connected, and projection of $S^1\times S^5$ onto $S^1$ gives its fundamental group: a based loop is a pair of coordinate loops; the second contracts because $S^5$ is simply connected, while the first lifts to $\mathbb{R}$ and its integer endpoint displacement classifies based homotopy. Choose its generator orientation below.

Choose small $3$-balls $D_i\subset C_i$ away from $A$ and paths in $C_i$ from their centres $u_i$ to $p$ or $q$, respectively. Fix an embedded arc $\alpha$ in $A$ from $p$ to $q$. A reference arc from $u_2$ to $u_1$ in the product, otherwise missing $A,C_1,C_2$, can be chosen in product coordinates; the loop obtained by adjoining the fixed sheet paths and $\alpha$ is null-homotopic since the product is simply connected. Replace a short segment of this reference arc by a detour through the connected-sum neck, around one generator of the $S^1$ factor, and back through the neck. Two parallel lanes make the outward and return portions disjoint. More formally, relative endpoint smoothing followed by the compact-arc embedding supplier gives an embedded representative of this path class; make its interior transverse to each of the three $3$-dimensional sheets, keeping short fixed endpoint collars normal to $C_i$. Since $1+3-6=-2$, its interior misses every sheet. Finitely many compactly supported perturbations suffice and preserve its relative path class and embeddedness. Denote the resulting embedded core arc by $\eta :u_2\to u_1$. By the construction, closing $\eta$ using the fixed sheet paths and $\alpha$ gives the generator $t$, not a null loop.

A sufficiently thin tubular neighbourhood of $\eta$ is $I\times D^5$: its normal bundle is trivial by projecting onto it in a Euclidean ambient embedding and transporting an initial basis by the skew matrix ODE $U^{\mathrm{\prime }}=[P^{\mathrm{\prime }},P]U$ along the interval. Choose a rank-$3$ subbundle in that normal bundle agreeing with the tangent $3$-planes of $C_i$ at its endpoints. Such a choice exists because the space of $3$-planes in ${\mathbb{R}}^5$ is path-connected; endpoint frames can be joined and interpolated on the interval. The resulting $I\times D^3$ has end balls $D_i$ after shrinking and straightening in endpoint charts. Remove their interiors from $C_1\cup C_2$ and insert the lateral cylinder $I\times S^2$, rounding its corners. Use the gluing that extends the specified orientations $C_1,-C_2$; an endpoint reflection realizes the required orientation convention. The tube and all rounding lie away from $A$ and the rest of the sheets. Each punctured $C_i$ is a $3$-ball, and two such balls joined by $S^2\times I$ form $S^3$. Thus the result is an embedded oriented sphere $B$; it agrees with $C_1$ near $p$ and with $-C_2$ near $q$. Their product tangent spaces are complementary to $TA$, so the only intersections are $p,q$ with signs $+1,-1$.

Take an embedded arc $\beta$ in $B$ from $q$ to $p$ running through the tube. Its part in the tube is homotopic relative endpoints to its core $\eta$ inside the tubular neighbourhood; its end parts are the fixed sheet paths up to homotopy in the punctured spheres. Consequently $[\alpha \ast \beta ]=t$ by step 2.1. Any other paths with the same endpoints in $A$ and $B$ are homotopic relative endpoints to these, since both sheets are $S^3$. In particular every admissible arc system gives the same nontrivial class. Normalize the label at $p$ to $1$; the label comparison lemma then gives the other label $t$, up to replacing the generator by its inverse under the opposite convention. The indices $1,-t$ are not negatives of each other, although their augmentation is zero. A disk filling a Whitney circle would contract $t$, impossible. Hence no Whitney disk or Whitney move exists for this pair.