What is the homology of the 1, 2 and 3 simplex? 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?
 A: 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.
