# An algebraic property of the simplicial boundary operator

I am working with simplicial homology with coefficients over $$\mathbb{Z}_2$$. Let $$C$$ be a null-homologous cycle in a simplicial complex $$K$$. Let $$\{C_i\}_{i=1}^n$$ be a basis for the cycle space of $$K$$ for some fixed dimension. We have $$C = \oplus_{i=1}^m C_{\ell_i}$$ for some subset $$\{C_{\ell_1},\dots,C_{\ell_m}\} \subseteq \{C_i\}_{i=1}^n$$ where $$\oplus$$ denotes the symmetric difference operator. Each $$C_{\ell_i}$$ is null-homologous, otherwise $$C$$ is not null-homologous. (This may not be completely accurate as there may be an even number of cycles from some non-trivial homology class in the formal sum making up $$C$$, however we can let their symmetric difference be one of the $$C_{\ell_i}$$'s which is null-homologous.)

Let $$\partial (F_{\ell_i}) = C_{\ell_i}$$ for each $$\ell_i$$, we have $$\partial (\oplus_{i=1}^m F_{\ell_i}) = C$$. Is the converse true? If $$\partial(F) = C$$ can we decompose $$F$$ into a formal sum $$F = \oplus_{i=1}^m F_{\ell_i}$$ such that $$\partial(F_{\ell_i}) = C_{\ell_i}$$?

First, I'm assuming that $$\mathbb{Z}_2$$ means the integers mod 2, not the 2-adic integers. Is that correct?
Am I allowed to choose the basis $$\{C_i\}$$? If so, the answer is yes. Maybe it's yes anyway, but you would need to clarify the part about "This may not be completely accurate ..." in the question.) Let me reorganize the setup, though. First, the vector space $$Z_d$$ of cycles in dimension $$d$$ has as a subspace $$B_d$$, the boundaries. Choose a splitting of the inclusion: write $$Z_d = B_d \oplus H_d$$ for some vector space $$H_d$$. Then choose a basis $$\{C_i\}$$ for $$B_d$$, and if you want, but I think it's not necessary, extend it to get a full basis of $$Z_d$$. Finally, in the entire chain group in dimension $$d$$, choose a complement $$A_d$$ for the space $$Z_d$$. Then the chain group in dimension $$d$$ will have the form $$A_d \oplus B_d \oplus H_d$$, and the boundary map will give an isomorphism $$A_{d+1} \to B_d$$: it's obviously surjective, and any element in the kernel would be in $$Z_{d+1}$$, the complement to $$A_{d+1}$$. Then if you consider the expression of any boundary as a symmetric difference of the basis elements for $$B_d$$, the answer to your question is yes, because this map is an isomorphism.
This should work over any field $$k$$ (replacing "symmetric difference" by "linear combination over $$k$$").