Here is the question I am trying to imagine and solve:
Compute the homology groups of the $\Delta$-simplex $X$ obtained from $\Delta^n$ by identifying all faces of the same dimension. Thus $X$ has a single $k$-simplex for each $k \leq n.$
I was told by my professor that:
The one simplex has $\mathbb Z$ (in odd dimension) and $\mathbb Z$ (in even dimension), the two simplex has $\mathbb Z$ (in odd dimension) and 0 and 0 and so on.
The 3 simplex has $\mathbb Z$ (in odd dimension) and 0 and 0 and $\mathbb Z$.
The 4 simplex has $\mathbb Z$ (in odd dimension) and all zeros.
The 5 simplex has $\mathbb Z$ (in odd dimension) and 0,0,0,0 and $\mathbb Z$.
I do not understand what my professor said and how he calculated the 3,4,5 simplices case for example, can someone explain this to me please?
By computations in the answers to Why if $n$ is odd the simplicial homology is $\mathbb Z,$ while if $n$ is even it is $0$?, for example, the chain complex looks like this: $$ \begin{eqnarray} \cdots & \xrightarrow{} & \mathbb{Z} & \xrightarrow{1} & \mathbb{Z} & \xrightarrow{0} & \mathbb{Z} & \xrightarrow{1} & \mathbb{Z} & \xrightarrow{0} & \mathbb{Z} & \\ & & 4 & & 3 & & 2 & & 1 & & 0 \end{eqnarray} $$ Of course you need to truncate this: all groups are zero in dimensions $n+1$ and higher if you're considering an $n$-dimensional simplex with the faces identified as in this problem. For example when $n=3$ you just have the chain complex $$ \begin{eqnarray} \cdots & \to & 0 & \xrightarrow{} & \mathbb{Z} & \xrightarrow{0} & \mathbb{Z} & \xrightarrow{1} & \mathbb{Z} & \xrightarrow{0} & \mathbb{Z} & \\ & & 4 & & 3 & & 2 & & 1 & & 0 \end{eqnarray} $$ Now compute homology. The kernel of the identity map (labeled "1" in these diagrams) is 0 while its image is all of $\mathbb{Z}$, and the kernel of the zero map is all of $\mathbb{Z}$ while its image is 0.
This should be enough information to compute the homology groups. What parts are still not clear? Start with $n=0$ (a point, homology should already be known) and then $n=1$ (a circle, as @ElliotYu said, and you should know the homology of a circle, and in any case, you can compute it from the chain complex $\mathbb{Z} \xrightarrow{0} \mathbb{Z}$), and then $n=2$ (not obviously any familiar space, but use the chain complex), and then $n=3$, until you see the pattern, both in the answer and in how the homology computation works in the chain complex.