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Let $X_g$ be the wedge of $g$ copies of the circle $S^1$ where $g>1$. Prove that $X_g \times X_g$ is not homotopy equivalent to a 3-manifold with boundary.

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Do you mind letting us know the source of this question? – Grumpy Parsnip Oct 6 '11 at 2:28
This question came around while studying the problem of classifying homotopy classes of maps $\sum_g \rightarrow X_g \times X_g$ which are onto on $\pi_1$ where $\sum_g$ is the surface of genus $g$. I guess a good place to start is by noting that $\pi_2(X_g \times X_g)=0$. – Manuel Oct 6 '11 at 4:21
up vote 4 down vote accepted

If it is a homotopy equivalent to a $3$-manifold $M$, looks at the homology long exact sequence for the pair $(M,\partial M)$ with $\mathbb Z_2$-coefficients. By Poincare duality, $H_i(M)\cong H_{3-i}(M,\partial M)$. You also know the homology groups of $M$, since you know those of $X$. If $\partial M$ has $c$ components, the fact that the long exact sequence has trivial euler characteristic allows you to compute that the rank of $H_1(\partial M;\mathbb Z_2)$ is equal to $2c+4g-2g^2-2<2c$. On the other hand, as you noted $\pi_2$ must be trivial. Yet any boundary component which is a sphere or $\mathbb{RP}^2$ will contribute some $\pi_2$, since it will represent a map of $S^2$ that doesn't bound a map of a $3$-ball to one side. Thus each boundary component has at least two homology generators, which is a contradiction.

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