I was given the following semisimplicial set.
$X_0 = \{v_1, v_2\}, X_1 = \{a, b, c_1, c_2\}, X_2 = \{s, t\}, d_0(a) = d_0(b) = v_2, d_1(a) = d_1(b) = v_1, d_0(c_1) = d_1(c_1) = v_1, d_0(c_2) = d_1(c_2) = v_2, d_{0,1,2}(s) = c_2, b, a, d_{0,1,2}(t) = b, a, c_1.$
Can you please help me understanding its fat geometric realization? I really don't understand how to glue things together in this specific case.
To compute the homology, I get $$\mathbb{Z}s \oplus \mathbb{Z}t \xrightarrow{\partial_2} \mathbb{Z}a \oplus \mathbb{Z}b \oplus \mathbb{Z}c_1 \oplus \mathbb{Z}c_2 \xrightarrow{\partial_1} \mathbb{Z}v_1 \oplus \mathbb{Z}v_2$$ with $\partial_1 = d_0 - d_1$ and $\partial_2= d_0-d_1+d_2$.
This is my first time doing this kind of exercise, so it is possibile that I make a lot of mistakes. We get that $\partial_1(c_1)=\partial_1(c_2)=\partial_1(a-b)=0$, so I would say that $\ker(\partial_1)=\mathbb{Z}c_1 \oplus \mathbb{Z}c_2 \oplus \mathbb{Z}(a-b)$ and $\text{im}(\partial_1)=\mathbb{Z}(v_2-v_1)$. In this way, $H_0(X, \mathbb{Z})=\mathbb{Z}v_1 \oplus \mathbb{Z}v_2 / \text{im}(\partial_1) \cong \mathbb{Z}$.
Then, $\partial_2(s)=c_2-b+a, \partial_2(t)=b-a+c_1$. Now, how do I proceed? Is $\ker(\partial_2) =0$ and $\text{im}(\partial_2) \cong \mathbb{Z}(c_2-b+a) \oplus \mathbb{Z}(b-a+c_1)$?
Thank you very much for your help.
We can have a visual conception of this space. It looks as below, but if you collapse the uppermost two vertices into one point. $s$ would be the top face and $t$ the bottom face, here. There is a fundamental subcircle giving this space a backbone and then some flaps left over; we can use an appropriate Mayer-Vietoris sequence based on this idea. Take a small thickening about that central circle on one hand, and the interiors of $s$ and $t$ one the other. Their intersection is a disjoint union of two contractible strips and we compute $H_\ast(X)\cong(\Bbb Z,\Bbb Z,0,0,\cdots)$ - basically the central circle is all that matters, homologically. This could also be a good exercise in cellular homology.
Your simplicial calculations all look correct to me, though I notice you did not actually calculate $H_2$ or $H_1$. $H_2$ will be zero because $\ker\partial_2=0$ and there is no "$\mathrm{im}\,\partial_3$" (well, there is, but it is also zero).
You should be able to check $H_1(X)\cong\Bbb Z$ from the work you've done so far. I might suggest re-writing $\Bbb Z(c_2-b+a)\oplus\Bbb Z(b-a+c_1)$ as something else first, though; have an explicit description of the elements of this subgroup.