TITLE: One question on cup product and torsion elements
QUESTION [10 upvotes]: As we know, by universal coefficient theorem, $H^{1}(X,\mathbb{Z})$ is torsion-free. My question is: 
for cup product $H^{1}(X,\mathbb{Z})\otimes H^{1}(X,\mathbb{Z})\rightarrow H^{2}(X,\mathbb{Z})$
could $a\cup b$ be a torsion element in $H^{2}(X,\mathbb{Z})$.

REPLY [3 votes]: You can also make an example involving a closed surface. Let $X$ be the connected sum of a torus $T$ and a projective plane, and let $f:X\to T$ be nontrivial on $H^2$. Two elements of $H^1(T)$ whose cup product is nonzero mod $2$ will pull back by $f$ to two elements of $H^1(X)$ whose cup product generates $H^2(X;\mathbb Z/2)=\mathbb Z/2$.