I want to show that $[\Delta^n_1-\Delta^n_2]$ is a generator of $H_n(\mathbb{S}^n)$ for $n \geq 1$, where $\Delta^n_i$ are identification of a small neighborhood of the upper hemisphere and the lower hemisphere with the standard simplex $\Delta^n$.

My attempt is to consider the Mayer-Vietoris sequence, which gives rise to the isomorphism $$\partial:H_n(\mathbb{S}^n) \to H_{n-1}(\mathbb{S}^{n-1})$$ where $\partial$ is the connecting homomorphism in the Mayer-Vietoris sequence.

Then I apply induction. I have proved the base case $n=1$. Now, consider $n=k$. My thought of using the inductive hypothesis is that since $\partial$ is an isomorphism, to show that $[\Delta^k_1-\Delta^k_2]$ is a generator of $H_k(\mathbb{S}^k)$, I need to show that $\partial[\Delta^k_1-\Delta^k_2]=[\Delta^{k-1}_1-\Delta^{k-1}_2]$. But I stuggle in doing so.

My question is: Is $\partial[\Delta^k_1-\Delta^k_2]=[\Delta^{k-1}_1-\Delta^{k-1}_2]$ true, and if yes, how can I show it? If not, is it possible that I can do it using the Mayer-Vietoris sequence argument above?

Thank you very much!

  • $\begingroup$ If you try writing it down explicitely for $n = 2$, then, you can compute what is $\partial[\Delta^2_1 - \Delta_2^k]$, and you will end up with a sum of three $1$-simplexes in $S^1$, you will have to glue two of them to get the cycle you are looking for. This kind of argument might be hard to generalise. Hatcher does it an other way, look at exemple 2.23 it Hatcher's Algebraic topology. $\endgroup$ – Robin Carlier Nov 1 '19 at 11:10
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    $\begingroup$ Your description of $\Delta^n_1$, $\Delta^n_2$ as "small neighborhoods" of the upper and lower hemisphere is not correct. Such neighborhoods must intersect to form a neighborhood of the equator, and hence $\Delta^n_1-\Delta^n_2$ would not even be an $n$-cycle. You should instead identify $\Delta^n_1$, $\Delta^n_2$ exactly with the closed upper and lower hemispheres, and you should also arrange that their boundaries correspond exactly. $\endgroup$ – Lee Mosher Nov 1 '19 at 13:06

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