= Solution
For a rational <homology (mathematics)> three-sphere and any <Spin-c structure>, the infinity version of <Heegaard Floer homology> is
$$
\boxed{HF^\infty(Y,\mathfrak s)\cong\mathbb Z[U,U^{-1}],\qquad\deg U=-2.}
$$
It is one Laurent tower, with one integral generator in each degree of a fixed parity and none in the opposite parity. Its absolute rational grading has an overall offset determined by $(Y,\mathfrak s)$.
To compute the <Euler characteristic> of the hat group, tensor with $\mathbb Q$, which preserves integral ranks. Put $V=HF^-(Y,\mathfrak s;\mathbb Q)$. It is a finitely generated graded $\mathbb Q[U]$ module. Localizing at $U$ gives $HF^\infty$, so the <structure theorem for finitely generated modules over a principal ideal domain> gives one free summand and finitely many $U$-primary torsion summands. The torsion is $U$-primary because it vanishes after localization, or directly because the reduced graded group is supported in finitely many degrees.
The exact sequence of <Heegaard Floer chain complexes> given by multiplication by $U$ and passage to the hat quotient identifies the hat group, as a parity-graded vector space, with an extension of $V/UV$ by $\ker(U:V\to V)$ with the latter parity reversed. The extra absolute shift by two between $CF^-/UCF^-$ and $\widehat{CF}$ is even, so it does not affect this argument. Thus
$$
\chi(\widehat{HF})=\chi(V/UV)-\chi(\ker U).
$$
The free summand contributes one vector to $V/UV$ and none to $\ker U$. A torsion summand $\mathbb Q[U]/U^r$, whose top generator has parity $\epsilon$, contributes one vector of parity $\epsilon$ to the quotient and one of the same parity to the kernel, since $U^{r-1}$ changes degree by an even number. Their contributions cancel. Only the single free summand remains, proving
$$
\boxed{\chi\bigl(\widehat{HF}(Y,\mathfrak s)\bigr)=(-1)^\epsilon\in\{1,-1\}.}
$$
Integer torsion does not enter an <Euler characteristic> defined by ranks.
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