Deducing inequality from exact triangle in Heegaard Floer homology

Question

In Hom's lectures on Heegaard Floer homology, pages 8 and 9 contain a proof that $rk \widehat{HF}(Y) \geq |H_1(Y, \mathbb{Z})|$ for rational homology spheres. The proof involves using an exact triangle and the statement that in rational homology spheres there a single free summand in $HF^{-}(Y, \mathfrak{s})$, for every torsion spin-$\mathbb{C}$ structure $\mathfrak{s}$. See below:

I understood that, since there is at least one free summand for each torsion spin-$\mathbb{C}$ structure, rk $HF^{-}(Y) \geq |H_1(Y, \mathbb{Z})|$. However, my question is, how does the exact triangle imply that $rk(HF^{-}(Y))=rk(\widehat{HF}(Y))$ (which would then make the inequality work)?

Answer 1

First, to fix a statement I made, the result in (2.2) is that there is not at least one, but exactly one free summand for each torsion spin-$\mathbb{C}$ structure in a rational homology sphere, causing $\text{dim}\;HF^{-}(Y)=|H_1(Y,\mathbb{Z})|$.

Now, because of the exact triangle we have $\text{Im}(/\langle U\rangle)\subset \widehat{HF}(Y,\mathfrak{s})$, which is the image of the southwest map from $HF^-(Y,\mathfrak{s})\to\widehat{HF}(Y,\mathfrak{s})$.

Now $\text{Im}(/\langle U\rangle)\cong \mathbb{F}_{(d)}\bigoplus_j\mathbb{F}_{(c_j)}$, so that $\text{dim}\;\widehat{HF}(Y,\mathfrak{s})\geq 1$, for every torsion spin-$\mathbb{C}$ structure (in fact, it's at least $1 + \#$torsion summands in $HF^-(Y,\mathfrak{s})$). Hence, $\text{dim}\;\widehat{HF}(Y)\geq \text{dim}\;HF^{-}(Y)=|H_1(Y,\mathbb{Z})|$ QED.