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.

[Expand]Notes for future reference - Alec (talk) 06:27, 1 February 2017 (UTC)

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

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.

[Expand]Notes to the marker: Alec (talk) 06:27, 1 February 2017 (UTC)

The proof for this exercise will either be transcluded, or copied and pasted into its own theorem or lemma page. As it falls under the Unified Mathematics project the proof is very methodical (for example indentation is directly related to the parse tree of a FOL statement) with small steps.

On paper this took me about half a page. However there's no guarantee that even I could read that; so I don't expect you to. Hence this.

I checked with the lecturer to be sure this qualified as "from the definitions", it seems to. He did mention that $\epsilon$ is usually used for the map I have called $I$ and mentioned some differences between the Abelian groups that usually have a $C$ in them and how he used $C$.

• This is annoying because I really don't like Hatcher's book, but adopted his notation here ($\Delta_n(X)$ - see bottom of page 104) to try and help the marker, rather than the more familiar notation involving $Z$s, $C$s) and $B$s

Note to future editors: the proof that the map, $I$, is surjective and actually a group homomorphism is omitted. Be sure to mark this as Template:Requires proof and mark it easy

Proof

Let us define some notation before we start.

• $X^n$ is the $n$-skeleton of $X$. Which made from all the simplices involved in $X$ of dimension $\le n$. So $X^1$ is itself a complex made up of all the 0 and 1 simplices.
• $X^{(n)}$ is not a complex but rather a set of all the $n$-dimensional simplices (and only those simplices) involved in $X$. For example $X^{(1)}$ are all the $1$-simplices of $X$ (and not the $0$-simplices), and so forth.
• $\Delta_n(X):\eq\mathcal{F}(X^{(n)})$ is the free abelian group with generators the set of $n$-simplices involved in $X$ ($\mathcal{F}(A)$ denotes the free abelian group generated by the elements $a\in A$)
• $\partial_n:\Delta_{n}(X)\rightarrow\Delta_{n-1}(X)$ is the boundary map.
  • $\partial_0:\Delta_0(X)\rightarrow 0$ is a group homomorphism onto the trivial group. $\partial_0:x\mapsto 0$ always; thus $\text{Ker}(\partial_0)\eq\Delta_0(X)$
• $$H_n^\Delta(X):\eq\frac{\text{Ker}(\partial_n)}{\text{Im}(\partial_{n+1})}$$

We also indulge in a few abuses of notation

• $c\in\Delta_0(X)$ means $c\eq\sum_\alpha n_\alpha v_\alpha$ where it is understood that $n_\alpha\in\mathbb{Z}$ is non-zero for only finitely many of the $\alpha$ in the implied indexing set. The indexing set for which each $\alpha$ is in can also be used to index $X^{(0)}$, thus $v_\alpha$ addresses each element of $X^{(0)}$ - under the identification of $v\in X^{(0)}$ with $1v\eq v\in\mathcal{F}(X^{(0)})$.
• $d\in\Delta_1(X)$ means $d\eq\sum_\alpha n_\alpha e_\alpha$ almost exactly as above but this time with $X^{(1)}$ instead of $X^{(0)}$ in play.
• For $e\in X^{(1)}$ (possibly identified with $1e\in\Delta_1(X)$) we use $e(0)$ for the initial point of the edge and $e(1)$ for the final point of the edge. Thus $\partial(e)\eq e(1)-e(0)$

We wish to compute:

• $$H_0^\Delta(X):\eq\frac{\text{Ker}(\partial_0)}{\text{Im}(\partial_1)}\eq\frac{\Delta_0(X)}{\text{Im}(\partial_1)}$$

Proof:

• Define $I:\Delta_0(X)\rightarrow\mathbb{Z}$ by $I:\sum_\alpha n_\alpha v_\alpha\mapsto \sum_\alpha n_\alpha$, clearly this is a group homomorphism and clearly also it is surjective (as $X$ is non-empty, we have at least one generator of $\Delta_0(X)$, so from that alone we can get any integer, that shows surjectivity. It being a group homomorphism is even easier).
  • By the first group isomorphism theorem we have: $$\frac{\Delta_0(X)}{\text{Ker}(I)}\cong_{\bar{I} }\mathbb{Z}$$ where $$\overline{I}:\frac{\Delta_0(X)}{\text{Ker}(I)}\rightarrow\mathbb{Z}$$ is the induced group isomorphism from $I$, thus $\overline{I}:[c]\mapsto I(C)$. This notation is unambiguous, $[c]$ represents an arbitrary equivalence class of the quotient (as this is first year work I will not elaborate any further. See the page first group isomorphism theorem for more information)
  • Suppose that $\text{Im}(\partial_1)\eq\text{Ker}(I)$. Then $\overline{I}$ would be an isomorphism from $H_0^\Delta(X)$ to $\mathbb{Z}$, and the result would be shown. This it the route we will take.
  • To do so we must show: $\text{Im}(\partial_1)\eq\text{Ker}(I)$. This will consist of two steps: $\text{Im}(\partial_1)\subseteq\text{Ker}(I)$ and $\text{Ker}(I)\subseteq \text{Im}(\partial_1)$
    1. Showing that $\text{Im}(\partial_1)\subseteq\text{Ker}(I)$, by the implies-subset relation, we need only show $\forall c\in\text{Im}(\partial_1)[c\in\text{Ker}(I)]$
       • Let $c\in\text{Im}(\partial_1)$ be given
         • By definition of image, $c\in\text{Im}(\partial_1)\iff \exists d\in \Delta_1(X)[\partial_1(d)\eq c]$
         • Thus let $d\in\Delta_1(X)$ be such that $\partial_1(d)\eq c$.
         • We observe now that $I(c)\eq I(\partial_1(d))$

           $\eq I(\partial_1(\sum_\alpha n_\alpha e_\alpha))$

           $\eq I(\sum_\alpha n_\alpha\partial_1(e_\alpha))$

           $\eq I(\sum_\alpha n_\alpha(e_\alpha(1)-e_\alpha(0))$

           $\eq I(\sum_\alpha n_\alpha e_\alpha(1) + \sum_\alpha (-n_\alpha) e_\alpha(0))$

           $\eq \sum_\alpha n_\alpha + \sum_\alpha (-n_\alpha)$

           $\eq \sum_\alpha(n_\alpha - n_\alpha)$

           $\eq 0$

         • Thus $I(c)\eq 0$, so $c\in\text{Ker}(I)$
       • Since $c\in\text{Im}(\partial_1)$ was arbitrary, we have shown that for all such $c$ that $c\in \text{Ker}(I)$. This completes the first step.
    2. Showing that $\text{Ker}(I)\subseteq \text{Im}(\partial_1)$. By the implies-subset relation we need only show $\forall c\in\text{Ker}(I)[c\in\text{Im}(\partial_1)]$
       • We require two lemmas before we can continue:
         1. $\forall v_0,v\in X^{(0)}\exists (p_i)_{i\eq 1}^k\subseteq X^{(1)}[\partial_1(\sum^k_{i\eq 1}p_i)\eq v-v_0\in\Delta_0(X)]$ (this requires path-connectedness) and
         2. $\forall c\in\Delta_0(X)\forall v_0\in X^{(0)}\exists d\in\Delta_1(X)[\partial_1(d)\eq c-I(c)v_0\in\Delta_0(X)]$
       • The proof of these can be found below (at this level of indentation)
         • Suppose the lemmas hold. We will prove the statement: $\forall c\in\text{Ker}(I)[c\in\text{Im}(\partial_1)]$
           • Let $c\in\text{Ker}(I)$ be given
             • We wish to show $c\in\text{Im}(\partial_1)$, by definition of image:
               • $c\in\text{Im}(\partial_1)\iff\exists d\in\Delta_1(X)[\partial_1(d)\eq c]$
             • Choose $v_0\in X^{(0)}$ arbitrarily
               • As $c\in\text{Ker}(I)\subseteq \Delta_0(X)$ we see $c\in\Delta_0(X)$, thus we can apply the second lemma
               • Choose $d\in\Delta_1(X)$ to be the $d$ posited to exist by the second lemma. So we have: $\partial_1(d)\eq c-I(c)v_0$
                 • now we wish to show our choice of $d$ is such that $\partial_1(d)\eq c$
                 • $\partial_1(d)\eq c-I(c)v_0$, we know this already
                   • But $c\in\text{Ker}(I)$, so $I(c)\eq 0$, thus
                 • $\partial_1(d)\eq c$
                 • As required.
             • Thus we have shown our choice of $d$ satisfies $\partial_1(d)\eq c$
           • Since $c\in\text{Ker}(I)$ was arbitrary we have shown $\forall c\in\text{Ker}(I)\exists d\in\Delta_1(X)[\partial_1(d)\eq c]$
             • Which as we established earlier is the same as $\forall c\in\text{Ker}(I)[c\in\text{Im}(\partial_1)]$ which was itself the same as $\text{Ker}(I)\subseteq\text{Im}(\partial_1)$
         • Thus we have shown that, provided the lemmas hold, $\text{Ker}(I)\subseteq\text{Im}(\partial_1)$
       • Proof of lemmas:
         1. Proof that: $\forall v_0,v\in X^{(0)}\exists (p_i)_{i\eq 1}^k\subseteq X^{(1)}[\partial_1(\sum^k_{i\eq 1}p_i)\eq v-v_0\in\Delta_0(X)]$
            • Let $v_0\in X^{(0)}$ be given
              • Let $v\in X^{(0)}$ be given
                • By path-connectedness of $X^1$ (the $1$-skeleton) we see there is a path through the $1$-skeleton, say $p$, from $v_0$ to $v$
                • Say $(p_i)_{i\eq 1}^k\subseteq X^{(1)}$ such that $p\eq\sum_{i\eq 1}^k p_i\in X^1$ which of course means $p\eq\sum_{i\eq 1}^k p_i\in\Delta_1(X)$ too
                • Thus: $\partial_1(\sum_{i\eq 1}^k p_i)\eq v-v_0\in\Delta_0(X)$ by the very definition of the path we took.
         2. $\forall c\in\Delta_0(X)\forall v_0\in X^{(0)}\exists d\in\Delta_1(X)[\partial_1(d)\eq c-I(c)v_0\in\Delta_0(X)]$
            • Let $c\in\Delta_0(X)$ be given, we shall write $c\eq\sum_\alpha n_\alpha v_\alpha$
              • Let $v_0\in X^{(0)}$ be given
                • We already know that for each $v_\alpha$ in $\sum_\alpha n_\alpha v_\alpha$ there exists a path $p(v_\alpha)$ say from $v_0$ to $v_\alpha$, $p(v_\alpha):\eq\sum_{i\eq 1}^k p_i$ such that $\partial_1(p(v_\alpha))\eq v_\alpha-v_0$ from the first lemma.
                • Choose $d:\eq\sum_\alpha n_\alpha p(v_\alpha)\in\Delta_1(X)$
                  • We must show that this $d$ satisfies the "such that" part of the lemma
                  • $\partial_1(d)\eq\partial_1(\sum_\alpha n_\alpha p(v_\alpha))$

                    $\eq \sum_\alpha n_\alpha \partial_1(p(v_\alpha))$

                    $\eq\sum_\alpha n_\alpha(v_\alpha - v_0)$

                    $\eq\sum_\alpha n_\alpha v_\alpha + (-\sum_\alpha v_\alpha) v_0$

                    $\eq c + (-I(c))v_0$

                    $\eq c - I(c)v_0$

                  • As required
                • Our choice of $d$ satisfies the requirements of the claim
              • Since $v_0\in X^{(0)}$ was arbitrary we have shown the claim for all such $v_0$
            • Since $v\in X^{(0)}$ was arbitrary we have shown the claim for all such $v$
  • We have now established $\text{Ker}(I)\eq\text{Im}(\partial_1)$ thus $H_0^\Delta(X)\cong\mathbb{Z}$ - as required.