Exercises:Saul - Algebraic Topology - 3/Exercise 3.2

Exercises

Exercise 3.2

Suppose that $(X,\mathcal{ J })$ is a non-empty path-connected topological space, equipped with a $\Delta$-complex structure. Show, directly from the definitions (Hatcher, of course...) that $H^\Delta_0(X)\cong\mathbb{Z}$

• We may assume without proof that the $1$-skeleton is path connected.

Notes

• $$H^\Delta_n(X):\eq\frac{\text{Ker}(\partial_0)}{\text{Im}(\partial_1)}$$

As $\partial_0$ just sends everything to $0$ we see $\text{Ker}(\partial_0)\eq \Delta_0(X)$ - all the vertices. Thus, essentially, $\text{Ker}(\partial_0)\cong\mathbb{Z}^{\#\text{vertices} }$.

Okay now $\partial_1:\Delta_1(X)\rightarrow\Delta_0(X)$, what is its image?

• First of all, for $f\in\Delta_1(X)$ we see $\partial_1(X):\eq\partial_1(\sum_{\alpha\in I_1}n_\alpha \sigma_\alpha)\eq\sum_{\alpha\in I_1}n_\alpha\partial_1(\sigma_\alpha)$ $\eq\sum_{\alpha\in I_1}n_\alpha(\text{Terminal}(\sigma_\alpha)-\text{Initial}(\sigma_\alpha))$
  • where $I_1$ is the set of $1$-simplices involved in $X$, and $n_\alpha\in\mathbb{Z}$ with $n_\alpha\neq 0$ for only finitely many of the $\alpha\in I_1$

This shows us that (sort of anyway) the image is spanned by various $\text{Terminal}(\sigma_\alpha)-\text{Initial}(\sigma_\alpha)$ (which are vertices)

Using $X^1$ to denote the $1$-skeleton (consistent notation be damned) then Caveat:and this is the informal part for any two vertexes of $X$, say $v_0$ and $v_1$, there is a path through the "edges" of $X^1$, such that $\partial_1(\text{that path})\eq v_1-v_0$

• We can make this a bit better! If $(p_i)_{i\eq 1}^k$ is a representation of the path, where $p_i$ is an edge, such that the initial vertex of $p_1$ is $v_0$ and the final vertex of $p_k$ is $v_1$ we can represent the path by the formal linear combination: $\sum_{i\eq 1}^k p_i$.
  • Note the $p_i$ need not be unique, it could have several loops in it, that wont matter (as the boundaries of cycles are $0$)

Now what we can sort of do is... well we want that sum to collapse to a few terms. What we can do is consider each $\sigma_\alpha$ which is an edge - plus the path from its endpoint to some fixed point of our choice. The boundaries then will have the same terminal point, subtracting various initials (and some/one (?) will have the same terminal and initial, giving us our "one less")

It would have to be shown this is in bijection with the edges.