# Torsion in $H^2(X,\mathbb{Z})$ induced by torsion in $H_1(X,\mathbb{Z})$

Let $X$ be a topological space. The universal coefficient theorem says that there is a short exact sequence $$0\rightarrow Ext(H_{k-1}(X,\mathbb{Z}),\mathbb{Z})\rightarrow H^{k}(X,\mathbb{Z})\rightarrow Hom(H_{k}(X,\mathbb{Z}),\mathbb{Z})\rightarrow 0$$ for $1\le k$. In particular a torsion element in $H_{1}(X,\mathbb{Z})$ induces a torsion element in $H^2(X,\mathbb{Z})$.

I would like to understand the statement above in the following situation. Consider a 2-torus (i.e. $(S^1)^2$) fibration $f:X\rightarrow Y$. Assume $H_{1}(Y,\mathbb{Z})=\pi_{1}(Y)=\mathbb{Z}/2\mathbb{Z}$ and monodromy representation is given by $(-id,-id)$ on $(S^1)^2$. Then the Spectral sequence associated with $f$ shows $$H_{1}(X,\mathbb{Z})=H_{1}(Y,\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}.$$ Is it possible to "see" the torsion in $H^2(X,\mathbb{Z})$ induced by $H_{1}(X,\mathbb{Z})$?

One may want to use Poincare duality (to realize the torsion as a submanifold etc), so please take $Y$ in your favorite dimension.

• I think that $X = RP^{2}$ is a simpler example to play around with. If you write out the cellular cochain complex for $RP^{2}$, then the algebraic reason for nonvanishing $H^{2}(X;\mathbb{Z})$ is pretty clear: The induced map on the cochain complex $H^{1}(X;\mathbb{Z})\to H^{2}(X;\mathbb{Z})$ is multiplication by 2. As for geometric intuition, I can't say I see it. – jmracek Dec 6 '12 at 16:12

## 1 Answer

Eh, have you looked at Hatcher's book? Specifically, the 2nd corollary to the universal coefficient thm.

• Whereas the the corollary cited may actually answer the question, self-contained answers are best. Something about what the citation says and possibly how it applies is very useful, especially for those who may not have access to the citation. – robjohn Dec 6 '12 at 16:16