0th Singular Homology

Warning: very boring post! Just revising some basic stuff in algebraic topology.

Consider a topological space $X$ and its connected components $X = \coprod_i X_i$ , can we say something about its singular homology?
Now, singular homology is built up (or perhaps quotiented down) from $S_\bullet (X) = \oplus_q S_q(X)$ , the module of singular chains. A $q$ -chain is a (formal) $\mathbb Z$ -linear combination of continuous maps $\sigma : \Delta_q \to X$ called singular $q$ -simplices (where $\Delta_q$ is the standard $q$ -simplex).

The fact that the standard $q$ -simplex is connected and that any singular simplex $\sigma$ is continuous tells us that the image must belong to exactly one connected component of the total space: $\sigma(\Delta_q) \subseteq X_i$ for some index $i$ .
Thus $S_\bullet (X) = \bigoplus_i S_\bullet(X_i)$ , (the sum is directed as the $S_\bullet(X_i)$ are disjoint by the same connectedness argument).

Homology’s final ingredient is the boundary map, $\partial_q : S_q(X) \to S_{q-1}(X)$ . Without giving a precise definition (which makes use of the orientation of the standard simplex), let’s just say that it sends a singular $q$ -simplex $\sigma$ to a certain linear combination of singular $(q-1)$ -simplices obtained by restricting $\sigma$ along the $(q-1)$ -faces of $\Delta_q$ , thus $\sigma \in S_q(X_i)$ implies $\partial_q(\sigma) \in S_{q-1}(X_i)$ .

Let’s call $\partial^i_q$ the restriction to the subspace $S_q(X_i) \subseteq S_q(X)$ , since these are disjoint we get $\partial_q = \sum_i \partial^i_q$ (with some healthy abuse of notation (?)) from which we get $B_q = {\rm Im} (\partial_{q+1}) = \bigoplus_i {\rm Im}(\partial^i_{q+1}) = \bigoplus_i B_q(X_i)$ .
Finally since $\sigma \in \ker(\partial_q) \iff \sigma \in \ker(\partial^i_q)$ for some $i$ , we get $Z_q(X) = \ker(\partial_q) \cong \bigoplus_i \ker(\partial^i_q) = \bigoplus_i Z_q(X_i)$ .

When quotienting out the borders we can distribute the direct sums finding $\begin{aligned} H_q(X) & = Z_q(X)/B_q(X) =\frac{\oplus_i Z_q(X_i)}{\oplus_j B_q(X_j)} \\ & = \bigoplus_i \frac{Z_q(X_i)}{\oplus_j B_q(X_j)} = \bigoplus_i Z_q(X_i)/B_q(X_i) = \bigoplus_i H_q(X_i) \end{aligned}$

Now let $X$ be an arc connected space and let $x_0 \in X$ .
A singular $0$ -chain is then a finite sum $c = \sum_k n_k \sigma_k$ , we can blurry the distinction between singular $0$ -simplices and points in the space, in fact let $x_k = \sigma_k$ .

Let $\gamma_k : I \to X$ be arcs connecting $x_0 = \gamma_k(0)$ to $x_k = \gamma_k(1)$ at its endpoints, since $I \cong \Delta_1$ we can interpret each arc as a singular $1$ -simplex!
Let’s calculate their images under the boundary map: $\partial_1(\gamma_k) = \gamma(1) -\gamma(0) = x_k - x_0$ .

Consider $b = \sum_k n_k\partial_1(\gamma_k)$ , it’s a border thus representing the zero class in homology, thus subtracting it from $c$ won’t change anything when we quotient out.

$c - b = \sum_k n_kx_k - \sum_j n_j(x_j - x_0)) = \sum_k n_k x_0 \in \mathbb Z \langle x_0 \rangle$ .
We’ve thus shown that, up to a border, any singular $0$ -chain is just a multiple of $x_0$ , so $H_0(X) \cong \mathbb Z\langle x_0 \rangle \cong \mathbb Z$ .

Going back to our original space $X = \coprod_i X_i$ we finally found

$H_0 (X) = \bigoplus_i H_0(X_i) \cong \bigoplus_i \mathbb Z$

in other words the 0th singular homology of a space is a free $\mathbb Z$ -module whose rank equals the number (or cardinality of the set of) its connected components!

Partially Ordered Thoughts

0th Singular Homology

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0th Singular Homology

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