First differential in cellular homology

Question

How do I compute the first differential of cellular homology in general for a CW complex $X$, without using for example that the first differential is zero if $X$ is path-connected and has one zero cell? (Question: doesn't path connectedness follow from having exactly one zero cell?)

By the definition in my lecture, fixing a one cell and a zero cell, we have to consider the map $$ S^0 \rightarrow X_0 \rightarrow X_0 / X_{-1} \cong \bigvee_{\text{$0$-cells}} S^0 \rightarrow S^0 \,. $$ The first map is the attaching map of the one cell, and the last map is the projection corresponding to the fixed zero cell in the wedge sum.

The isomorphism to the wedge sum comes in higher dimensions from identifications $D^k / S^{k-1} \cong S^{k-1}$. But in our case we would have $D^0 / S^{-1} = \mathrm{pt} \neq S^0$. Then it isn't clear how the map $\mathrm{pt} \rightarrow S^0$ would look like, so I cannot determine the degree of the total map $S^0 \rightarrow S^0$.

I don't see how this definition is working out for the first differential or what I can do instead to compute it directly, for example in the case of a $1$-sphere with two $0$-cells and two $1$-cells.

Answer 1

First, we have $S^{-1} = \emptyset$, and by convention (in this context) $X / \emptyset \cong X_+$, i.e., $X$ with a disjoint point added, for any space $X$. Hence, we have $D^0 / S^{-1} \cong S^0$, which is consistent with the same fact in higher dimensions.

Second, there is an easy way to read off the differential of $1$-cell in (cellular) homology. In terms of the attaching map $\varphi: S^0 \to X^0$ of a $1$-cell (edge) $e$, write $S^0 \cong \{1, -1\}$. Then, $$de = \varphi(1) - \varphi(-1).$$ In other words, the differential of a $1$-cell is just the signed sum of its two boundary $0$-cells.

Answer 2

The cellular chain complex of a CW-complex $X$ with $n$-skeleta $X^n$ is usually defined by $C^{CW}_n(X) = H_n(X^n,X^{n-1})$ and $\partial^{CW}_n : C_n(X) = H_n(X^n,X^{n-1}) \stackrel{\partial_n}{\to} H_{n-1}(X^{n-1}) \stackrel{j_*}{\to} H_{n-1}(X^{n-1},X^{n-2}) = C_{n-1}^{CW}(X)$.

It is easy to see that we can naturally identify $C_n(X)$ with the free abelian group whose generators are the $n$-cells of $X$. Note, however, that the proof is slightly different for $n > 0$ and $n = 0$. For $n > 0$ we use the fact that $H_n(X^n,X^{n-1}) \approx H_n(X^n/X^{n-1},*) = H_n(X^n/X^{n-1})$ and observe that $X^n/X^{n-1}$ is a wedge of $n$-spheres, one for each $n$-cell of $X$. For $n = 0$ we can of course use the convention $Y/\emptyset = Y^+$ = disjoint union of $Y$ with a one-point space $*$ which gives $H_0(X^0,X^{-1}) \approx H_n(X^n/\emptyset,*) = H_n((X^0)^+,*) \approx \tilde H_0((X^0)^+)$. Formally we can of course regard $(X^0)^+$ as wedge of $0$-spheres with base point $*$, one for each $0$-cell of $X$, which allows to conclude that $H_0(X^0, X^{-1}) \approx \tilde H_0((X^0)^+)$ is a free abelian group whose generators are in $1$-$1$-correspondence with the $0$-cells of $X$. But I think this point of view obscures the simple fact that $H_0(X^0, X^{-1}) = H_0(X^0)$ = free abelian group generated by the point of $X^0$ which is discrete space.

Hatcher (in his book "Algebraic Topology") writes

Next we describe how the cellular boundary maps $\partial^{CW}_n$ can be computed. When $n = 1$ this is easy since the boundary map $\partial^{CW}_1 : H_1(X^1,X^0) \to H_0(X^0)$ is the same as the simplicial boundary map $Δ_1(X) \to Δ_0(X)$. In case $X$ is connected and has only one $0$ cell, then $\partial^{CW}_1$ must be $0$, otherwise $H_0(X)$ would not be $\mathbb Z$. When $n > 1$ we will show that $\partial^{CW}_n$ can be computed in terms of degrees:
Cellular Boundary Formula. [ ... ]

Note that Hatcher introduces the simplicial chain complex $\Delta_*(X)$ for so-called $\Delta$-complexes $X$ (see the section "Simplicial Homology"). The group $\Delta_n(X)$ is the free abelian group generated by the (open) $n$-simplices $e^n_\alpha$ of $X$; alternatively one can take as generators the characteristic maps $\sigma^n_\alpha : \Delta^n \to X$ of the $e^n_\alpha$ which are very special singular simplices.

$\Delta$-complexes are less general than CW-complexes, but the $1$-skeleton $X^1$ of a CW-complex $X$ is always a $\Delta$-complex. Although Hatcher claims it is easy to see what $\partial^{CW}_1 : H_1(X^1,X^0) \to H_0(X^0)$ looks like, he does not give a rigorous description of $\partial^{CW}_1$. The essential thing is that we need a separate argument differing from the case of $\partial^{CW}_n$ for $n > 1$.

It is well-know that if $(X,A)$ is a pair of topological spaces and $\sigma^n : \Delta^n \to X$ is a singular simplex such that $\sigma^n(\partial \Delta^n) \subset A$, then

1. $\partial_n \sigma^n \in C_{n-1}(A)$ so that the equivalence class $\langle \sigma^n \rangle \in C_n(X,A) = C_n(X)/C_n(A)$ is a cycle,

2. $\partial_n ([\langle \sigma^n \rangle ]) = [\partial_n \sigma^n] \in H_{n-1}(A)$, where $[ - ]$ denotes homology class.

The same is true in the simplical chain complexes of a pair $(X,A)$, where $X$ is a $\Delta$-complex and $A$ is a subcomplex (but in that case we only consider characteristic maps $\sigma^n_\alpha)$).

Let us now see what we can extract from Hatcher's book. For $n > 1$ Hatcher considers the diagram on p. 141. For $n = 1$ we do not have such a diagram including reduced homology groups. Instead we get the simple

$\require{AMScd}$ \begin{CD} H_1(D^1/S^0) @<{q_*}<{\approx}< H_1(D^1,S^0) @>{\partial_1}>> H_0(S^0)\\ @V{i_{\alpha*}}VV @V{\Phi_{\alpha*}}VV @VV{\varphi_{\alpha*}}V \\ H_1(X^1/X^0) @<{q_*}<{\approx}< H_1(X^1,X^0) @>{\partial_1}>> H_0(X^0) \end{CD}

The generator $g_\alpha$ of $H_1(X^1,X^0)$ corresponding to the $1$-cell $e^1_\alpha$ has the form $g_\alpha = \Phi_{\alpha*}(g)$, where $g$ is the canonical generator of $H_1(D^1,S^0)$. It therefore suffices to understand $\partial_1 : H_1(D^1,S^0) \to H_0(S^0)$.

In Theorem 2.27 Hatcher proves that if $X$ is a $\Delta$-complex and $A \subset X$ is a subcomplex, then the canonical chain map $Δ_n(X,A)→C_n(X,A)$ sending each $n$-simplex $e^n_\alpha$ of $X$ to its characteristic map $σ^n_\alpha :Δ^n→X$ induces an isomorphism $\iota_n : H^Δ_n(X,A)→H_n(X,A)$.

$D^1$ is a $\Delta$-complex with two $0$-simplices $e^0_\pm = \{\pm 1\}$ and one open $1$-simplex $e^1 = (-1,1)$. As the characteristic map for $e^1$ we take the linear homeomorphism $\chi^1 : \Delta^1 \to D^1$ mapping $v_0$ to $-1$ and $v_1$ to $1$. Clearly $S^0$ is a subcomplex of $D^1$.

$\Delta_1(D^1)$ is the free abelian group generated by $\sigma^1$ and $\Delta_1(S^0) = 0$. Hence $\Delta_1(D^1, S^0) = \Delta_1(D^1)/0 = \Delta_1(D^1)$. Since $\chi^1(\partial \Delta^1) \subset S^0$. we see that $\chi^1$ is a cycle in $\Delta_1(D^1, S^0)$. Moreover $\Delta_2(D^1) = \Delta_2(S^0) = 0$, thus $\Delta_2(D^1,S^0) = 0$ whence $H_1^\Delta(D^1,S^0) = \Delta_1(D^1)$. Therefore $\chi^1$ is the canonical generator of $H_1^\Delta(D^1,S^0)$. We conclude that $H_1(D^1,S^0)$ has $\iota_1(\chi^1) = [\langle \chi^1 \rangle]$ as its canonical generator. By 1. and 2. above we have $$\partial_1([\langle \chi^1 \rangle]) = [\partial_1 \chi^1] = [\chi^1(v_1) - \chi^1(v_0)] = (+1) - (-1).$$ Recall that $H_0(S^0)$ is generated by its two points $\pm 1$.

Our above diagram shows that the generator $g_\alpha$ of $H_1(X^1,X^0)$ corresponding to the $1$-cell $e^1_\alpha$ with attaching map $\varphi_\alpha : S^0 \to X^0$ is mapped to $H_0(X^0)$ via $$\partial_1(g_\alpha) = \varphi_\alpha(+1) - \varphi_\alpha(-1) .$$ Note that $H_0(X^0)$ is generated by the $0$-cells of $X$.

Update:

It is unnecessary to use the simplicial homology groups of $\Delta$-complexes to prove that $[\langle \chi^1 \rangle]$ is the canonical generator of $H_1(D^1,S^0)$.

We know that $H_1(D^1,S^0) \approx H_1(D^1/S^0) \approx H_1(S^1) \approx \mathbb Z$. From the long exact sequence of the pair $(D^1,S^0)$ we get the exact sequence $$0 = H_1(D^1) \to H_1(D^1,S^0) \stackrel{\partial_1}{\to} H_0(S^0) $$ which shows that $\partial_1$ is injective. We have shown that $\partial_1([\langle \chi^1 \rangle]) = (+1) - (-1)$ using only 1. and 2. above (we did not use the $\Delta$-complex structure of $D^1$). But the element $(+1) - (-1) \in H_0(S^0) = \mathbb Z\langle +1\rangle \oplus \mathbb Z\langle -1\rangle$ is not a multiple of any other element of $\mathbb Z\langle +1\rangle \oplus \mathbb Z\langle -1\rangle$, thus it must be a generator of the infinite cyclic subgroup $\operatorname{im}(\partial_1) \subset \mathbb Z\langle +1\rangle \oplus \mathbb Z\langle -1\rangle$. This is only possible if $[\langle \chi^1 \rangle]$ generates $H_1(D^1,S^0)$.