Why $C_i(X)=\mathbb{Z}$ for all $i$ in the proof below?

Question

If $X$ is a point, then $H_n(X)=0$ for $n>0$.

I understand the part that $H_0(X)\simeq\mathbb{Z}$.

What confuses me a lot is:

In the proof below, how can we have $\mathbb{Z}$ for all the chain groups $C_i(X)$? If $X$ is a point, then shouldn't it be only $C_0(X)=\mathbb{Z}$ and $C_i(X)=0$ for $i>0$? In other words, a point is a 0-simplex, how can we have 1-simplex, 2-simplex and so on since we don't have a line, a 2-dimensional ball etc..

Another of my doubt is why exactly $\partial(\sigma_n)$ is zero when $n$ is odd and $\sigma_{n-1}$ when $n$ is even? And why can there be some isomorphism boundary maps?

I have been spending too much time on this proof. Could someone please give some light?

Thanks.

Answer 1

Look at some simple cases the only 1 simplex is $[a,a]$ where $a$ is your point. So $$\partial [a,a]=a-a=0.$$ Now look at $2$ simlplex, $[a,a,a]$ $$\partial [a,a,a]=a-a+a=a$$. So you get isomorphisms alternating with the zero map.

Answer 2

By definition, a singular $n$-simplex in $X$ is a continuous map $\Delta^n\to X$. If $X=\{x\}$ has one point, then there is exactly one such map for any $n$, namely the map $\sigma_n:\Delta^n\to X$ given by $\sigma_n(y)=x$ for all $y$.

By definition, the boundary of a singular $n$-simplex $\Delta^n\to X$ is an alternating sum of its compositions with certain maps $\Delta^{n-1}\to\Delta^n$ (the inclusions of each of the faces of $\Delta^n$). In our case, all of those compositions must be the same, since there is only one possible map $\Delta^{n-1}\to X$, namely $\sigma_{n-1}$. So we have $$\partial\sigma_n=(-1)^0\sigma_{n-1}+(-1)^1\sigma_{n-1}+\dots+(-1)^n\sigma_{n-1}.$$ Every other term in this sum cancels, so if there are an even number of terms you get $0$ and if there are an odd number of terms you can cancel all but the first. Since there are $n+1$ terms, this says $$\partial\sigma_n=0$$ if $n$ is odd and $$\partial\sigma_n=(-1)^0\sigma_{n-1}=\sigma_{n-1}$$ if $n$ is even.