Instead of the regular definition of the homology group using free abelian groups on $n$-simplices, could we instead define it using symmetric difference?

For example, let $X$ be a topological space.

  1. For each $n$, let $G_n$ be the group of $n$-submanifolds without boundary with the operation $A \cdot B = A \Delta B \setminus \partial (A \Delta B)$.
  2. Let $\partial_n : G_n \to G_{n - 1}$ be the homomorphism mapping manifolds to their boundary.
  3. Let $H_n = \ker \partial_n / \text{im}\ \partial_{n + 1}$.
  • $\begingroup$ Have you tried your proposal out on any examples? How about taking $X = S^1$? $\endgroup$
    – Rob Arthan
    Feb 24 '20 at 1:13
  • $\begingroup$ It's weird that you are focusing on the symmetric difference, when the primary difference between your definition and the usual one is that you are using submanifolds instead of simplices. That's a much more drastic change. (Talking about symmetric differences pretty much amounts to just using mod $2$ coefficients instead of integer coefficients.) $\endgroup$ Feb 24 '20 at 1:37
  • $\begingroup$ There are also various obvious problems with your proposal as written. What do you mean by "submanifold" if $X$ is just a topological space? Also, your boundary map is just identically $0$ since you said you are only using manifolds without boundary. $\endgroup$ Feb 24 '20 at 1:39
  • $\begingroup$ By manifolds without boundary, I mean that the manifolds don't contain their boundary. For example, for a sphere, $G_2$ would contain open disks. $\endgroup$
    – user623070
    Feb 24 '20 at 1:41

Your group operation runs into serious set theoretic difficulties, even for elementary examples.

For example, suppose $X = \mathbb R^2$, $A$ is the circle of radius $3$ centered on $(-1,0)$, and $B$ is the circle of radius $3$ centered on $(+1,0)$. How does your formula for $A \cdot B$ define a $1$-submanifold without boundary? The situation could be even worse, it is possible for to find even smooth circles $A,B$ embedded in $\mathbb R^n$ such that $A \cap B$ is a very complicated subset, even a Cantor set.

Early topologists ran into similar difficulties when they tried to formalize the concept of a "path" in a topological space $X$. For example, defining "path" as "subspace of $X$ homeomorphic to $[0,1]$" leads to all kinds of difficulties: if I walk along a path from $p \in X$ to $q \in X$, and if I walk along a path from $q \in X$ to $r \in X$, is the result that I have walked along a path from $p$ to $r$? Unfortunately no, because the union of two subspaces homeomorphic to $[0,1]$ is not a subspace homeomorphic to $[0,1]$.

What turned out to work very nicely was to formalize a path as a kind of function, namely a continuous function $f : [0,1] \to X$. Now it's true to say that if I walk along a path $f : [0,1] \to X$ from $p=f(0)$ to $q=f(1)$, and if I walk along a path $g : [0,1]$ from $q = g(0)$ to $r=g(1)$, then walking along the first path (at half speed) and then along the second path (at half speed) results in a path from $p$ to $r$. This is just the ordinary formula from the theory of fundamental groups for concatenation of two paths.

In short, the reason for constructing homology theory by basing it on formal sums of simplices is that the theory is easy. It gets around all sorts of technical, set theoretic difficulties that arise not just in your own formalization but many other formalizations.


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