In the exercises classes of my algebraic topology course we often informally use that transforming a cycle continuosly doesn't change is class in singular homology. My lecturer told me, that it is probably impossible to make this intuition precise in singular homology, because simplexes do not look nice or regular at all in the theory.
But I have read the answer to this question
Homotopic vs. homologous simplices/chains
(which is about simplicial homology), and believe that it generalizes to singular homology. My question: Is the following statement true?
Statement. Suppose $X$ is a topological space and $\sum_in_i\,c_i$ is a cycle in $C^{sing}_m(X)$. Suppose there are homtopies $h^i: \Delta^m\times I\to X$ from $c_i$ to $a_i$ which satisfy the following condition:
\begin{align*}
c_i\circ d_k = c_j \circ d_l \qquad \Rightarrow \qquad \forall t: \, h^i_t\circ d_k = h^j_t \circ d_l
\end{align*}
Then $\sum_i n_i\, a_i$ is a cycle in $C^{sing}_m(X)$ homologous to $\sum_in_i\,c_i$.
Remark. $d_k$ are the face maps $d_k: \Delta^{m-1} \to \Delta^m$ leaving out the $k$ -th coordinate. They are $d_k = [e_0,e_1,...,\hat e_k,,...,e_m]$ in Hatcher's notation. The boundary of a cycle $\sum_in_i\,c_i$ is \begin{align*} \partial \left(\sum_in_i\, c_i\right) = \sum_i \sum_{k=0}^m n_i\, (-1)^k\, c_i \circ d_k = 0\,. \end{align*} The coefficients $n_i$ are integers.
Proof sketch Without loss of generality assume $n_i=+1,-1$. If not, repeat some of the $c_i$'s and their homotopies in the sum. Let $q: \bigsqcup_i \Delta^n_i \to K$ be the equalizer of the two maps \begin{align*} \bigsqcup_{c_rd_k = c_jd_l}\Delta^{m-1} \xrightarrow{\bigsqcup\, \text{inc}_rd_k} \bigsqcup_i\Delta^{m}_i\end{align*} and \begin{align*} \bigsqcup_{c_rd_k = c_jd_l}\Delta^{m-1} \xrightarrow{\bigsqcup\, \text{inc}_jd_l} \bigsqcup_i\Delta^{m}_i\end{align*} Here $\text{inc}_r$ is the $r$-th embedding $\Delta^m \hookrightarrow \bigsqcup_i\Delta_i^m$. Set $x_i = q\circ \text{inc}_i$ and consider the chain $x = \sum_in_ix_i$ in $C_m^{sing}(K)$. Check that it is a cycle.
$\bigsqcup_i h^i_t$ equalizes the two maps above for each $t$ and thus induces a map $h_t: K \to X$. It follows from a theorem by Whitehead that the induced map $h: K \times I \to X$ is continuous. Check that $(h_0)_*(x) = \sum_i n_ic_i$ and $(h_1)_*(x) = \sum_in_ia_i$. The result follows since singular homology is homotopy invariant.
Edit: I've just relized that
Question about Singular Homology section in Hatcher
is relevant (the corresponding paragraph in Hatcher respectively). The space $K$ in the proof is the space $K_\xi$ in Hatcher's book.
