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The Lickorish-Wallace theorem tells us that any closed 3-manifold $Y$ is an integer link surgery on $S^3$, which yields an oriented cobordism between $S^3$ and $Y$. Filling out the $S^3$ by a 4-ball $B^4$, we obtain a compact 4-manifold bounding $Y$, and as a corollary we have that the 3rd oriented cobordism group $\Omega^{SO}_3=0$.

Now think of spin$^c$ analogue of this situation. We start with a spin$^c$ 3-manifold $(Y,\mathfrak{t})$. The Lickorish-Wallace theorem applied to the underlying manifold $Y$ gives a cobordism $W$ which only consists of 2-handles. To fill out the $S^3$-side of $W$ by a 4-ball, we should impose the condition that the $S^3$ is endowed with the (unique) torsion spin$^c$ structure $\mathfrak{s}_0$. Thus, a temporary version for spin$^c$ Lickorish-Wallace theorem is the existence of a spin$^c$ structure on $W$ which extends $\mathfrak{t}$ and restricts to the torsion spin$^c$ structure on $S^3$. But this cannot be true in full generality, as already seen from the case when $W$ is the product cobordism. (In this case, $\mathfrak{t}$ should be the torsion spin$^c$ structure.)

But (it seems) it is a forklore that a sufficient condition to make the theorem work is requiring that $\mathfrak{t}$ is torsion. For clarity, let me state the modified theorem again:

For a link surgery cobordism $W$ from $(S^3,\mathfrak{s}_0)$ to a closed 3-manifold $Y$ and a torsion spin$^c$ structure $\mathfrak{t}$ on $Y$, we can always find a spin$^c$ structure which restricts to $\mathfrak{s}_0$ on $S^3$ and $\mathfrak{t}$ on $Y$ (so that $(Y,\mathfrak{t})$ represent 0 in $\Omega^{Spin^c}_3$).

Question. How we can prove this theorem? This theorem appears in many places, e.g. it is a key step in defining absolute $\mathbb{Q}$-gradings in Heegaard Floer homology. But those literature do not include the proof. I guess a few obstruction theory would work, but it is mysterious to me how to exploit the torsion condition to extend the spin$^c$ structure.

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First, let me remark that $S^3$ has a unique spin$^c$ structure $\mathfrak{t}_0$, which is also torsion. (I decided to call it $\mathfrak{t}_0$, because I prefer to use $\mathfrak{t}$ for spin$^c$ structures on 3-manifolds, and $\mathfrak{s}$ for spin$^c$ structures on 4-manifolds.) So, you don't need to specify that $\mathfrak{t}_0$ is the unique torsion spin$^c$ structure, and this probably will remove the source of your confusion.

In general, spin$^c$ structures on 3- and 4-manifolds are an affine space (a torsor, some say) over $H^2$; more precisely, for a 3-manifold $Y$ (respectively, a 4-manifold $W$) there is a free and transitive action of $H^2(Y;\mathbb{Z})$ on $\rm{Spin}^c(Y)$ (resp., of $H^2(W;\mathbb{Z})$ on $\rm{Spin}^c(W)$)). I'll denote the action with a + (probably not the best choice, but it works).

Moreover, the action is compatible with the restriction map of spin$^c$ and the restriction map on cohomology. That is, if $\iota: Y \hookrightarrow W$, $\alpha\in H^2(W;\mathbb{Z})$, and $\mathfrak{s}$ is a spin$^c$ structure on $W$, then $\iota^*\alpha + \iota^*\mathfrak{s} = \iota^*(\alpha + \mathfrak{s})$.

In your case, $W$ is a 2-handlebody, so $\iota^*: H^2(W;\mathbb{Z}) \to H^2(Y;\mathbb{Z})$ is onto, since the next term in the long exact sequence for the pair is $H^3(W,Y;\mathbb{Z}) \cong H_1(W,S^3) \cong H_1(W) = 0$, where the first equality is Poincaré-Lefschetz duality, the second is excision, and the third is the absence of 1-handles.

Combining the action and the surjectivity of $\iota^*: H^2(W;\mathbb{Z}) \to H^2(Y;\mathbb{Z})$, you get that every spin$^c$ structure $\mathfrak{t}$ on $Y$ extends to some spin$^c$ structure on $W$. (The restriction on the $S^3$ boundary is unique, so there is no problem there.)

This actually proves the stronger version of the statement that you wanted, without the restriction on being torsion. (So your confusion was, in a sense, justified.)

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  • $\begingroup$ Thanks for the correction and clear answer. For some reason I confused $S^3$ with $S^1\times S^2$, both of which appear often in Heegaard Floer theory. $\endgroup$
    – cjackal
    Apr 16, 2018 at 21:17

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