Let $X$ be a topological space. The singular homology groups are the homology groups which arise from the singular chain complex. This is a chain complex

$\dots \longrightarrow C_k(X) \longrightarrow C_{k-1}(X) \longrightarrow \dots$

where each $C_k(X)$ is generated by the singular $k$-chains. A singular $k$-chain is a topological mapping from the standard $k$-simplex into $X$.

From this definition it is evident that a singular $k$-chain can be, well, singular. For example, it may map a high-dimensional simplex onto a single point.

What if we restrict the class of chains? I define $C_k^{\rm reg}(X)$ as the $\mathbb Z$-module generated by the topological embeddings of the standard simplex into $X$. When we consider the chain complex

$\dots \longrightarrow C_k^{\rm reg}(X) \longrightarrow C_{k-1}^{\rm reg}(X) \longrightarrow \dots$

then again we can compute the homology groups.

Question: How do these homology groups relate? Under which conditions is the embedding of the second complex into the first complex an isomorphism on homology?


This is a much more difficult chain complex to work with. If say $X$ is an $m$-manifold, then there will not be any "regular" simplices of dimension $>m$. Therefore in the "regular" homology, every $m$-cycle will give a distinct homology class. So say $H_1^{\text{reg}}(S^1)$ will be uncountable. We can divide up the usual generator of the homology of $S^1$ into a sum of two regular 1-chains in uncountably many ways and these will give distinct elements of $H_1^{\text{reg}}(S^1)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.