# Why not use symmetric difference to define homology?

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 $$H_n = \ker \partial_n / \text{im}\ \partial_{n + 1}$$.
• Have you tried your proposal out on any examples? How about taking $X = S^1$? Feb 24 '20 at 1:13
• 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.) Feb 24 '20 at 1:37
• 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. Feb 24 '20 at 1:39
• 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.
– user623070
Feb 24 '20 at 1:41

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.