Reduced homology

Question

I'm trying to understand why $H_0(X) \cong \tilde{H}_0(X) \oplus \mathbb{Z}$. In Hatcher on page 110 he writes

''...Since $\varepsilon \partial_1 = 0$, $\varepsilon$ vanishes on $im \partial_1$ and hence induces a map $H_0(X) \rightarrow \mathbb{Z}$ with kernel $\tilde{H}_0(X)$...''

The bit I don't understand is how exactly it induces this map. I mean, where does $\mathbb{Z}$ come from all of a sudden. It's in the extended diagram for the reduced homology $\tilde{H}_0$ and the $H_0(X)$ is in the regular diagram. So somehow there is a map from normal homology into reduced homology (the exact sequences), but how? Many thanks for your help.

Answer 1

There is a map $\epsilon : C_0(X) \to \mathbb Z$. It is indeed a part of the complex for reduced homology, but it exists independently. Because $\epsilon \partial_1$ is zero, the value of $\epsilon(z)$ doesn't change when you add a boundary so, by definition, it defines a map $\overline \epsilon : H_0(X) = C_0(X)/\mathrm{im} \partial_1 \to \mathbb Z$.

Now, what is the kernel of this map? It's the set of (nonreduced) homology classes represented by an element on which $\epsilon$ vanishes. So that's $\ker \epsilon / \mathrm{im} \partial_1$. That's the very definition of $\widetilde H_0(X)$.

There is really nothing deep going there (and nothing too interesting either). Reduced (co)homology is nothing but a very handy convention that avoids too much boring distinction between zeroth and positive degree for computations you do everyday (looking at the homology of contractible space, looking at the homology of a wedge sum...)

Answer 2

Conceptually, the key thing to digest is this fundamental fact:

• In homology, every cycle, considered as an abstract space in its own right, is a boundary. Except 0-cycles. For 0-cycles to be boundaries, you need a balanced number of $+$ and $-$ signs.

So the point of reduced homology is to create a context where all cycles are boundaries (in possibly larger spaces -- say if you're dealing with the space $X$, all cycles in $X$ are boundaries in the cone on $X$, $CX$. But that's just one way to make sense of this). So then the problem of whether or not a cycle in a fixed space is a boundary becomes an extension problem. The cycle may be an abstract boundary, but the object that it's the boundary of may not map into the space you want.

This problem becomes more elaborate in other homology theories, like singular bordism. Because here you have many manifolds which themselves are not boundaries -- so even constant maps from these manifolds to your space $X$ represent non-trivial elements of singular bordism. So they don't tell you anything interesting about your space $X$.

But your technical question has a technical answer. You have your augmentation map $H_0(X) \to \mathbb Z$ given by taking the sums of the signs of your 0-cells, mapped into $X$. The kernel of this map is reduced homology. But any map to $\mathbb Z$ is split, so there is a map back $H_0(X) \to \overline{H_0}(X)$. This map back I believe is not natural unless you know $X$ is connected.

Answer 3

First, the map $\epsilon$ defined previously in the text is a map $C_0(X) \to \mathbb{Z}$. Then after verifying that $\epsilon \partial_1 = 0$ (use the definition of $\epsilon$ to do this), we know that $\epsilon$ maps anything in the image of $\partial_1$ to $0$, hence factors through $C_0(X)/ im \partial_1$, which by definition is the group $H_0(X)$.

Now what is perhaps not clear on first reading is that $\tilde{H}_0(X)$ is simply defined to be the kernel of the induced map $H_0(X) \to \mathbb{Z}$.

Hope this helps!

Answer 4

Perhaps this is implicit in the above answers, but $\mathbb{Z}$ comes from the fact that $H_*(pt.)=H_0(pt.)=\mathbb{Z}$. Everything can be seen as coming from the map unique $X \to pt.$ and looking at the induced map in homology. Then $\tilde{H}_*(X)$ is the kernel of the induced map in homology. You are just getting rid of the contribution of the basepoint. This seems pointless now perhaps, but later on it will be helpful to have both. Some long exact sequences are easier to understand when you don't need to worry about non-zero $H_0$'s, like computing the homology of $S^n$ inductively using the Mayer-Vietoris Sequence.

Answer 5

How about using first and third isomorphism theorems of groups?

We have $\text{Im} \;\partial_1 \le \text{Ker}\; \epsilon \le \text{Ker}\; \partial_0=C_0(X)$ and both $\text{Im}\; \partial_1$ and $\text{Ker}\;\epsilon$ are normal subgroups of $C_0(X)$. Thus by third isomorphism theorem of groups, $\cfrac {\cfrac{C_0(X)}{\text{Im} \;\partial_1}}{\cfrac{\text{Ker} \;\epsilon}{\text{Im}\; \partial_1}} \approx \cfrac {C_0(X)}{\text{Ker} \;\epsilon}.$

But since $\epsilon$ is surjective, we have $\cfrac {C_0(X)}{\text{Ker} \;\epsilon} \approx \Bbb Z$ by first isomorphism theorem of groups.

By definition of singular homology, we have $H_0(X)=\cfrac {C_0(X)}{\text{Im} \; \partial_1}.$ From here it comes naturally that $\tilde{H}_0(X)$ must be defined as $\cfrac{\text{Ker} \; \epsilon}{\text{Im} \;\partial_1}.$

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EDIT: $\text{Im} \;\partial_1 \le \text{Ker}\; \epsilon$ comes from $\epsilon\partial_1=0$.