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[I needed to remember how homology works.. see picture 1]

Recall that in simplicial homology we have: $C_1(X) = \mathbb{Z}\left[e_1, e_2, e_3\right]$ and $C_2(X) = \mathbb{Z}
\left[f_{12},f_{23},f_{31}\right]$ with the boundary operator $\partial:C_2 \to C_1(X)$ given by
$$
\partial(f_{12}) = e_2 - e_1\\
\partial(f_{23}) = e_3 - e_2\\
\partial(f_{31}) = e_1 - e_3
$$
and note that $\partial(f_{12} + f_{23} + f_{31}) = 0$. In other words, we have that $\partial(f_{12})+ \partial(f_{23}) = - \partial(f_{31})$. This tells us that $H_1(X;\mathbb{Z})$ is generated by (the equivalence class) of $\partial(f_{31})$ in $C_1(X)$. In other words, we have $H_1(X;\mathbb{Z}) \cong \mathbb{Z}$.

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[Now we can talk about the "multiplication by two" map on the circle.. see picture 2 and 3]

Let $Z$ be the circle with six vertices and consider the map $f: Z \to X$ given by
$$
f(a_1) = f(a_4) = e_1\\
f(a_2) = f(a_5) = e_2\\
f(a_3) = f(a_6) = e_3
$$

Recall that the cochain complex are given by taking the "hom-set" functor and the differential are given by the pullbacks:

$d: \text{Hom}(C_n(X);\mathbb{Z}) \to \text{Hom}(C_{n+1}(X);\mathbb{Z})$

where given a map $f: C_n(X) \to \mathbb{Z}$, we have $d(f) = f \circ \partial$ given by $C_{n+1} \xrightarrow{\partial} C_{n} \xrightarrow{f} \mathbb{Z}$.

Mon, 05 Aug 2019 07:00 GMT