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 $n$-torus $(S^1)^n$-fibration $f:X\rightarrow Y$. Assume that $H\_{1}(Y,\mathbb{Z})=\pi\_{1}(Y)=\mathbb{Z}/2\mathbb{Z}$ and monodromy representation is given by $(-id,\dots,-id)$ on $(S^1)^n$. Then the Spectral sequence associated with $f$ shows
$$
H\_{1}(X,\mathbb{Z})\cong H\_{1}(Y,\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}.
$$ 
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 your favorite $Y$ (satisfying the condition above or something similar) and $n\in \mathbb{N}$. 


Thank you for your assistance. 



Torsion in $H^2(X,\mathbb{Z})$ induced by torsion in $H_1(X,\mathbb{Z})$
Having the cycles and boundaries of degree one, it's possible to construct a cohomology class in $H^2(X)$ that corresponds to your torsion element. This is explained in the following: 


**1. Homological algebra**


Let $(C,d)$ be a chain complex of free abelian groups, $G$ an abelian group and let 
$$\varphi: Ext(H\_{q-1}(C),G) \to H^q(C,G)$$ be the monomorphism from UCT. An inspection of 
the proof in [Spanier: Algebraic Topology, Theorem 5.5.3] reveals: Set $Z\_q = \ker(d\_g)$, $B\_{q-1}=\text{im}(d\_q)$. Then there is an exact sequence 
$$\text{Hom}(Z\_{q-1},G) \to \text{Hom}(B\_{q-1},G) \to Ext(H\_{q-1}(C),G) \to 0.\hspace{50pt}(\ast)$$



> 
> If $e \in Ext(H\_{q-1}(C),G)$ is represented by $f: B\_{q-1} \to G\;$ then 
>  $\varphi(e) \in H^q(C,G)$ is represented by the cycle $f \circ d\_q: C\_q \to G$. 
> 
> 
> 


**2. The finitely generated case**


For simplicity let's assume each $C\_q$ has finite rank. Let $\alpha \in H\_{q-1}(C)=Z\_{q-1}/B\_{q-1}$ 
be a generator of a direct summand of order $n$. Choose a representative $x \in Z\_{q-1}$ 
of $\alpha$. We can find decompositions $Z\_{q-1} = Z' \bigoplus \mathbb{Z}x,\;B\_{q-1}=B' \oplus \mathbb{Z}nx$ with $B' \le Z'$. Let $f: B\_{q-1} \to \mathbb{Z},\;f|B'=0,\;f(nx)=1$. Then $f$ represents $e \in Ext(H\_{q-1},\mathbb{Z})$ via $(\ast)$ and $e$ corresponds to $\alpha$ under the non-canonical isomorphism $Ext(H\_{q-1}(C),\mathbb{Z}) \cong H\_{q-1}(C)\_{tor}$. By 1., $h=f \circ d\_q: C\_q \to \mathbb{Z}$ is a cocycle that represents $\varphi(e)$. 


**3. An example**


Let $X$ be the CW-complex in [exercise 4](http://www.stanford.edu/class/math215b/Sol2.pdf) with $\Pi\_1(X)=Q\_8$ (quaternion group) and let $C=C(X)$ be the cellular chain complex. $X$ has 0-cells $x, y$, 1-cells $a,b,c,d$, 2-cells $p, q, r$ and one 3-cell. The boundary operator is given by: 
$$d\_1(a)=d\_1(b)=d\_1(c)=d\_1(d)=x-y\hspace{130pt}$$
$$d\_2(p)=a-b+c-d,\;\;d\_2(q)=c-b+d-a,\;\;d\_2(r)=d-b+a-c$$
Set $A = a-b,\; B = a-c,\;C=a-d$. Then $Z\_1 = \langle A,B,C\rangle$ and $B\_1 = \langle A-B+C,2B,2C \rangle$. 


Hence $B$ represents a homology class in $H\_1(X)$ of order 2. Define $f: B\_1 \to \mathbb{Z}$ by $$f(A-B+C)=0,\; f(2B)=1,\; f(2C)=0.$$ Now by 2., the cocycle $h=f \circ d\_2$: 
$$h(p)=0,\;h(q)=0,\;h(r)=1$$
represents a cohomology class in $H^2(X)$ of order 2. 


**Remark:** Note that we didn't need to consider $d\_3: C\_3 \to C\_2$ in order to construct our torsion class in $H^2(X)$. This, in contrast, would be necessary if we computed $H^2(X)$ directly. 


