Let $\sigma_1,\sigma_2 \in C_1(X)$ be two singular $1$-simplices that are cycles in a space $X$. If $\sigma_1$ and $\sigma_2$ are homotopic is it true that the equivalance classes $[\sigma_1]$ and $ [\sigma_2]$ are equal in $H_1(X)$?
Asked
Active
Viewed 109 times
1
-
Near-duplicate: https://math.stackexchange.com/questions/1483953/homotopic-vs-homologous-simplices-chains – Eric Wofsey Feb 03 '20 at 22:40
-
The question does not make sense. In general a singular $1$-simplex $\sigma$ is not a cycle, thus $[\sigma]$ is not defined. – Paul Frost Feb 03 '20 at 23:06
-
Well, he does assume that each of $\sigma_1,\sigma_2$ is a cycle. – Lee Mosher Feb 04 '20 at 02:45
-
@LeeMosher You are right, I didn't read carefully. – Paul Frost Feb 05 '20 at 18:37
1 Answers
4
No. Take a homotopy which shrinks $\sigma_1$ to a constant map, i.e $H(x,t) = \sigma_1(xt)$ (making the identification $\Delta^1 = [0, 1]$).
A constant 1-simplex is a cycle and is the boundary of the constant 2-simplex, this would mean that $[\sigma_1] = [\text{constant simlpex}] = 0$ but the space $S^1$ for example is a counterexample to this since its first homology is generated by a singular 1-simplex and is nontrivial.
Your mistake was not requiring the homotopy $H$ between $\sigma_1 \text{ and } \sigma_2$ to satisfy the property of $H(-,t)$ being a cycle for all t. If it is a cycle for all $t$ then what you say is correct.
This can be proven as a consqeuence of the "homotopy invariance" of homology: If $f$ and $g$ are homotopic maps $Z \rightarrow X$ then $f_* = g_* : H_*(Z) \rightarrow H_*(X)$.
Noel Lundström
- 5,152