= Solution
The map is a homogeneous <Laurent polynomial ring> homomorphism between the two rank-one infinity towers. Choose homogeneous generators $e_i$ with absolute degrees $q_i$. Its degree is
$$
\Delta(W,\mathfrak s)=\frac{c_1(\mathfrak s)^2-2\chi(W)-3\sigma(W)}4.
$$
Consequently the general graded-module description is
$$
\boxed{F^\infty_{W,\mathfrak s}(e_1)=a\,U^r e_2,\qquad a\in\mathbb Z,\qquad r=\frac{q_2-q_1-\Delta(W,\mathfrak s)}2.}
$$
If the grading does not allow an integer exponent, the map is zero. Negative exponents are allowed in the infinity theory. The notation $F$ here denotes the map called $\Phi^\infty$ in the question.
The standard topological statements for <Heegaard Floer cobordism maps>, which require no proof in this part, are
$$
\boxed{b_2^+(W)>0\ \Longrightarrow\ F^\infty_{W,\mathfrak s}=0,\qquad b_2^+(W)=b_1(W)=0\ \Longrightarrow\ F^\infty_{W,\mathfrak s}\text{ is an isomorphism}.}
$$
In the second case $a=\pm1$ and the map is therefore $\pm U^r$ with the exponent determined by the displayed degree formula. Since both boundary components are rational <homology (mathematics)> <spheres>, the rational <intersection form> is nondegenerate, so $b_2^+=0$ means that it is negative definite, allowing also the rank-zero form.
The printed hypothesis $b_1(Y_i)=0$ does not imply $b_1(W)=0$. Thus one must not replace the second implication by an unconditional isomorphism assertion for every negative-definite <cobordism>. A concrete check is $W=(S^1\times S^3)\setminus(B^4\sqcup B^4):S^3\to S^3$. It has $b_2^+=0$ and $b_1(W)=1$. A <one-handle> followed by a <three-handle> gives $e\mapsto e\otimes\theta^+$ and then kills the $\theta^+$ summand; hence its infinity map is zero. Its grading shift is one, also incompatible with a nonzero map between the even-graded $S^3$ towers. With only the stated boundary assumption, the general homogeneous Laurent-module description and the two topological implications above are the valid conclusions; determining any remaining coefficient requires the actual <cobordism>, not just its boundary Betti numbers.
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