In Proposition 4 on page 6 of this paper, http://arxiv.org/abs/1202.6302, the authors claim to produce a degree 2 $\pi_1$-surjective map $f$ between $M=S^1 \times \Sigma_2$ and $N=\#_2 S^2 \times S^1$ (the connected sum of two copies of S^2 \times S^1). I cannot follow their construction and I seem to have quickly convinced myself that such a map is impossible. The fundamental group of $N$ is a free group $\mathbb{Z} \ast \mathbb{Z}$ which has no non-trivial center. Letting $h$ be the generator of $\pi_1(S^1)$, this implies that $f_*(h \times 1)= 1$. Consider the abelianized map $$ f_* : H_1(M) \to H_1(N)$$ Then $f_*([h \times 1])=0$. Which means that the image of $$ f^*:H^1(N) \to H^1(M)$$ vanish on $[h \times 1]$. Now write the generator $ \alpha_3 \in H^3(N)$ as a product of elements $\alpha_1 \cup \alpha_2$ with $\alpha_i \in H^i(N)$. However we then obtain that $f^*(\alpha_3)=0=f^*(\alpha_1) \cup f^*(\alpha_2)=0$. To see this observe that $f^*(\alpha_1)$ vanishes on $[h \times 1]$ and thus under the identification of $H^1(S^1\times \Sigma_2)=\mathbb{Z}[h \times 1]^* \oplus H^1(\Sigma_2)$ has no component in $\mathbb{Z}[h \times 1]^*$...
EDIT: As Jim Conant and Igor Belegradek pointed out, my error is in the deduction that the cup-product must be zero just because this component vanishes. $f^*(\alpha_1)$ could be a generator in $[1 \times \pi_1(\Sigma_2)]^*$ and f^*(\alpha_2) could be of the form $[h \times 1]^* \otimes H^1(\Sigma_2)$. Thanks very much to them.
Does this make sense or am I making a mental error?
Moreover any subgroup of a free group is free and I think this would imply that there is no map between the two manifolds of non-trivial degree. What am I missing?
