Give example(s) of elements that inside torsion part of cohomology group after quotient

Question

For a manifold $X$, its (integer-valued) cohomology group $H^k(X,\mathbb Z)=Z^k(X,\mathbb Z)/B^k(X,\mathbb Z)$$=\text{Ker}_{\partial_k^*}/\text{Im}_{\partial^*_{k-1}}~$ is construct from the chain complex: $$ \cdots\mathop\longrightarrow^{\partial^*_{k-1}} C^k(X,\mathbb Z)\mathop\longrightarrow^{\partial^*_k}C^{k+1}(X,\mathbb Z)\mathop\longrightarrow^{\partial^*_{k+1}}C^{k+2}(X,\mathbb Z)\mathop\longrightarrow^{\partial^*_{k+2}}\cdots $$ where $ C^k(X,\mathbb Z)=\text{Hom}_{\mathbb z}(C_k(X,\mathbb Z),\mathbb Z)~ $ is the group of all $\mathbb Z$-linear maps acts on singular k-chains.
We know that the torsion $ T^k $ part of $ H^k(X,\mathbb Z) $ is equal to $ T_{k-1} $, the torsion part of $H_{k-1}(X,\mathbb Z)$
Also $ f\in Z^k(X,\mathbb Z)\Leftrightarrow f|_{B_k(X,\mathbb Z)}=0 $, $ g\in B^k(X,\mathbb Z)\Leftrightarrow g|_{Z_k(X,\mathbb Z)}=0 $, thus when talking about elements of $H^k(X,\mathbb Z)$, there is no need to consider how an element of $ Z^k(X,\mathbb Z) $ acts on non-closed singular k-chains(this info will be divided, if I'm not wrong here). After this step, I failed to construct torsion inside $H^k(X,\mathbb Z)$.
I wonder what the elements of its torsion part look like in $ Z^k(X,\mathbb Z) $ as maps before quotient, such as the elements of $ H^2(\mathbb{RP}^2,\mathbb Z)\simeq \mathbb Z_2 $. There might need several maps in different equivalent classes of $H^k(X,\mathbb Z)$ to show its torsion.

Answer 1

Again, solved the problem myself...
The problem is: The action info of cochains on non-closed chains is not fully divided in terms of the cohomology group.
For example, if a non-closed singular k-simplex $\sigma$ has a boundary $\partial \sigma = nc^{k-1}$, $n\in\mathbb Z$, $c^{k-1}\in Z_{k-1}(X)$, though a closed cochain $f\in Z^k(X)$ can acting on the simplex whatever it want, a exact cochain $g\in B^k(X)$ is slightly different. There must exist a $k-1$ cochain $h\in C^{k-1}(X)$, such that $g(\sigma^k)=\partial^*h(\sigma)=h(\partial\sigma)$ $=h(nc)=n~h(c)$. If $\partial\sigma$ is decompose properly, $h(c)$ has no constraint. The only problem is n, if $|n|>1$, it can no longer span the ray of all chains $\mathbb Z\sigma^*$ restrict on this simplex $\sigma$. ($\sigma^*(\sigma)=1$), thus the torsion group $\mathbb Z_n=\mathbb Z/n\mathbb Z$ appears.
Consider a more concrete example: $X=\mathbb{RP}^2$. It has the CW structure $\sigma^2\cup\sigma^1\cup\sigma^0$, where $\partial\sigma^2=2\sigma^1$, $\partial\sigma^1=0$, $\partial\sigma^0=0$. Its $Z^2(X)$'s element could simplified as a map $f:\{\sigma^2\}\to\mathbb Z$, because the 3rd chain group $C_3(X)$ is $0$, thus no constraint on this map. But focus on $B^2(X)$, $g(\sigma^2)=\partial^*h(\sigma^2)=2h(\sigma^1)$, it can only maps $\sigma^2$ to a even number. thus $B^2(X)=\text{Hom}_{\mathbb Z}(\sigma^2,2\mathbb Z)\simeq 2\mathbb Z$. We can thus get the cohomology group $H^2(X)=Z^2(X)/B^2(X)=\mathbb Z/2\mathbb Z=\mathbb Z_2$, fits the inference of universal coefficient theorem.