# Do two elements in the same homology class have the same homology?

Do two elements in the same homology class have the same homology?

Specifically, let's say we have a topological space $X$ and two singular $n$-cycles which are also generators of the chain group of $X$, i.e. $\sigma_1, \sigma_2: \Delta_n \to X$ continuous and $\sigma_1, \sigma_2 \in Z_n(X)$.

Now assume that it is true that $[\sigma_1]=[\sigma_2] \in H_n(X)$; then is it the case that all of the homology groups of $\sigma_1(X)$ are equal to those of $\sigma_2(X)$ (when $\sigma_1(\Delta_n) \subseteq X$ and $\sigma_2(\Delta_n) \subseteq X$ have the subspace topology)?

I think the answer is no, because the zeroth and first homologies of $S^1$ and $S^1 \sqcup S^1$ are different, since for the first $H_0(S^1)=H_1(S^1)=\mathbb{Z}$ and for the second $H_0(S^1 \sqcup S^1)=H_1(S^1 \sqcup S^1) = \mathbb{Z} \oplus \mathbb{Z}$, but when their embeddings in $\mathbb{R}^3$ are considered as 1-cycles, I think that they might be homologous (as $1-$cycles in $Z_1(\mathbb{R}^3)$) because there is a cobordism between them, thus their oriented difference should be a boundary.

See Figure 1 here. This question is basically about the pair of pants cobordism. See also my previous question on MathOverflow.

• How do you define "$\sigma(X)$" (recalling that $\sigma$ is a map from $\Delta^n$ to $X$)? Do you mean the image of $\sigma$? (Also, cobordisms are pretty irrelevant here.) Nov 20, 2016 at 12:21
• @NajibIdrissi Sorry for the typo. Also, would you mind explaining how cobordisms are "pretty irrelevant here"? Nov 20, 2016 at 13:12
• Well being homologous is much more general than being cobordant (it's not even trivial that cobordant submanifolds will have homologous fundamental classes in the ambient manifold by the way), and if you have a random singular $1$-simplex its image is not guaranteed to be a manifold. Besides $H_1(\mathbb{R}^3) = 0$... Any $1$-cycle in $\mathbb{R}^3$ is homologous to any other one. Nov 20, 2016 at 13:21
• If you have two disjoint compact oriented submanifolds $M, M' \subset N$ and an embedded cobordism $W \subset N$ between $M$ and $M'$, then the pushforwards of the fundamental classes $[M]$ and $[M']$ will be homologous in $H_n(N)$, because manifolds can be triangulated. Nov 20, 2016 at 13:27
• It's simpler than that, actually. Then the fundamental class $[W] \in H_n(W, \partial W)$ gives a chain in $C_{n+1}(N)$ whose differential is, basically by definition, $d[W] = [M'] - [M]$ (and actually you don't need triangulations I think, I made a mistake). Nov 20, 2016 at 19:29

A circle and figure eight in the plane are both images of 1-cycles (homologous since $H_1(\mathbb{R}^2)=0$), but have different $H_1$.