# Traditional homology is a special case of reduced homology

Published on 8 August 2017

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## Is this new mathematics?

No. Just an argument over mathematical presentation, bringing together individual observations previously made by e.g. Hatcher 2002.

## What mathematical presentation?

It is standard practice to define ‘reduced’ homology after and on the basis of ‘traditional’ homology. But reduced homology is actually more fundamental, while traditional homology can be seen as secondary. Accordingly, I prefer to write $$H_i(X)$$ for the $$i$$th reduced homology group, and $$H_i(X)_{\emptyset}$$ for the $$i$$th traditional homology group of a topological space $$X$$.

## Traditional and reduced homology are almost identical.

Yes, for any non-empty space $$X$$, the difference between traditional and reduced homology boils down to $$H_0(X)_{\emptyset} \simeq H_0(X) \times \mb{Z}$$, so there’s not that much at stake here.

## What kind of homology are we talking about?

Singular homology. (The arguments for cohomology are analogous.)

## In what way is reduced homology more fundamental?

Reduced homology has a more parsimonious interpretation, it has a more natural definition, and traditional homology can in turn be elegantly defined in terms of reduced homology.

## A more parsimonious interpretation?

Reduced homology is a better measure of the number of 1-dimensional holes in a space.

## The 0th traditional Betti number encodes the number of connected components of a space. So the 0th reduced Betti number encodes “the number of connected components minus 1”. How is that more parsimonious?

Isn’t it unfortunate that with traditional homology, a contractible space, in particular a point — a simple space with no holes whatsoever — nonetheless has non-trivial homology in degree $$0$$? Isn’t it strange that whereas the $$n$$th Betti number generally represents the number of $$n+1$$-dimensional ‘holes’ in a structure, for $$n = 0$$, it should be the number of connected components? With reduced homology, contractible spaces have trivial homology and the 0th Betti number actually encodes the number of 1-dimensional holes: the number of ‘gaps’ between components.

## How do you define the number of ‘gaps’ between components? If a topological space $$X$$ has $$k$$ components, isn’t every pair of components separated by a gap, for a total of $$k!/2$$ gaps if $$k \geq 2$$? Whereas the 0th reduced Betty number is $$k - 1$$?

For 1-dimensional holes, we need a 1-dimensional perspective. This we get by projecting the topological space onto a line, in which case we find exactly $$k-1$$ gaps between its components.

## How can reduced homology be more fundamental if it assigns trivial homology both to contractible spaces and to the empty space?

The reduced homology of the empty space isn’t generally discussed. But if you follow the definition, you find that its $$-1$$st reduced homology group is $$\mb{Z}$$.

## That just sounds wrong.

It is exactly what we want if homology encodes holes. The empty space has a hole of a unique, 0-dimensional kind: it is empty. In general, the prototypical space with a single $$n+1$$-dimensional hole is the $$n$$-sphere. The empty space is the $$-1$$-sphere (Definition 1) and it has the reduced homology of a sphere (Figure 1).

Definition 1: Let $$n \geq -1$$ and let $$\left\|-\right\|$$ be the euclidean norm on $$\mathbb{R}^{n+1}$$. The $$n$$-sphere is the topological subspace $$S^n = \{a \in \mathbb{R}^{n+1} | \left\|a\right\| = 1\}$$. The $$n+1$$-disk is the topological subspace $$D_{n+1} = \{a \in \mathbb{R}^{n+1} | \left\|a\right\| \leq 1\}$$.

\begin{array} {lccccccc} & H_{-1} & H_0 & H_1 & H_2 & H_3 & H_4 & \dots\\ S^{-1} & \mb{Z} & 0 & 0 & 0 & 0 & 0 & \dots\\ S^0 & 0 & \mb{Z} & 0 & 0 & 0 & 0 & \dots\\ S^1 & 0 & 0 & \mb{Z} & 0 & 0 & 0 & \dots\\ S^2 & 0 & 0 & 0 & \mb{Z} & 0 & 0 & \dots\\ S^3 & 0 & 0 & 0 & 0 & \mb{Z} & 0 & \dots\\ S^4 & 0 & 0 & 0 & 0 & 0 & \mb{Z} & \dots\\ \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \end{array}

Figure 1: The groups $$H_i(S^n)$$

## Is defining this empty $$-1$$-sphere in any way useful?

This is a bit tangential, but there exists an elegant proof that establishes the homology groups of the $$n$$-sphere through induction on $$n$$, starting at $$n = -1$$. In other words, we can derive the homology of the spheres from the homology of the empty space. Without providing full details, the proof, as covered in Hatcher 2002, hinges on the following theorem:

Theorem 1: For any topological space $$X$$, and any two subspaces $$A, B \sbeq X$$ such that $$A \cup B = X$$, there exists a short exact sequence

$$0 \lra A \cap B \lra A \oplus B \lra X \lra 0$$

It is possible to apply this to $$X = S^n$$, for any $$n \geq 0$$, by taking $$A$$ and $$B$$ to be two overlapping hemispheres of $$X$$ homeomorphic to the disk $$D_n$$, such that $$A \cap B$$ is homeomorphic to $$S^{n-1}$$:

$$0 \lra S^{n-1} \lra D_{n} \oplus D_{n} \lra S^n \lra 0$$

The Meyer-Vietoris theorem applied to this short exact sequence gives us a long exact sequence of homology groups. Because disks are contractible, the terms $$H_i(D_n \oplus D_n)$$ are all trivial, hence $$H_{i+1}(S^{n+1})$$ and $$H_{i}(S^{n})$$ are isomorphic for all $$i$$ and all $$n \geq -1$$. We can thus derive any homology group of any sphere through induction from the homology groups of $$S^{-1}$$, the empty space: for any $$n \geq -1$$, all homology groups of $$S^n$$ are trivial, except $$H_{n}(S^n)$$, which is isomorphic to $$\mb{Z}$$.

With traditional homology, this derivation is arguably less elegant because then induction cannot start with the empty space and the term $$H_0(D_n \oplus D_n)$$ is non-trivial.

## You said that reduced homology has a more elegant definition than traditional homology. How is this possible given that reduced singular homology is usually defined by taking the singular chain complex of a space $$X$$, from the definition of traditional homology, and extending it with a special map $$\epsilon: C_0(X) \lra \mb{Z}$$.

That construction is unnecessarily complicated. There is a more natural definition of reduced homology, and of $$\epsilon$$ in particular. We obtain $$\epsilon$$ automatically as part of the chain complex if we remove an arbitrary restriction from the definition of traditional singular homology. Just watch the definitions unfold:

Definition 2: For any $$n \geq -1$$, the standard $$n$$-simplex $$\Delta_n$$ is the $$n$$-simplex spanned by the standard basis $$(e_0,e_1,\dots,e_n)$$ of $$\mathbb{R}^{n+1}$$, i.e. the set

$$\left\{(w_0,w_1,\dots,w_n) \in \mathbb{R}^{n+1} \Bigg| \sum_{0 \leq i \leq n}w_i = 1, w_0,w_1,\dots,w_n \geq 0\right\}.$$
A singular $$n$$-simplex $$\sigma$$ in a topological space $$X$$ is a continuous map $$\Delta_n \longrightarrow X$$. For any $$n \geq i \geq -1$$, let $$\iota_i: \Delta_n \lra \Delta_{n+1}$$ be the linear map which sends any $$e_j$$ to $$e_j$$ if $$j \leq i$$ and to $$e_{j+1}$$ if $$j < i$$.

Definition 3: Define $$C: \mf{Top} \lra \mf{Ch_{\bullet}(Ab)}$$ by letting $$C_n: \mathfrak{Top} \lra \mathfrak{Ab}$$ be the composition of the functor $$\Cont(\Delta^n,-): \mf{Top} \lra \mf{Set}$$ which sends a topological space $$X$$ to the set of all singular $$n$$-simplices in $$X$$ and the free functor $$\mf{Set} \lra \mf{Ab}$$, and by having $$d_n: C_n(X) \lra C_{n-1}(X)$$ send any generating $$n$$-simplex $$\sigma$$ of $$C_n(X)$$ to $$\sum (-1)^{i} \sigma \circ \iota_i$$, for every $$n \in \mathbb{Z}$$. Then for every $$n \in \mb{Z}$$, the $$n$$th homology functor

$$H_n: \mf{Top} \lra \mf{Ab}$$
is the composition of $$C$$ with the homology functor $$H_n: \mf{Ch_{\bullet}(Ab)} \lra \mf{Ab}$$.

## Isn’t this just the definition of traditional singular homology?

If we wanted to obtain traditional homology, we would have to restrict $$n \geq 0$$ in Definition 2. But there is no intrinsic reason for this restriction, as the case $$n = -1$$ is still well-defined: $$\Delta^{-1} = \emptyset$$, therefore $$\Cont(\Delta^{-1},X)$$ is a singleton for any space $$X$$, and consequently, $$C_{-1}(X) \simeq \mb{Z}$$ and $$d_0 = \epsilon$$ maps every $$0$$-simplex in $$C_{0}(X)$$ onto the unique $$-1$$-simplex generating $$C_{-1}(X)$$. In contrast, the restriction $$n \geq -1$$ is really necessary, because $$\mathbb{R}^{n+1}$$ is not defined for $$n < -1$$.

Also, note in particular that since $$C_0(\emptyset) = 0$$, we get $$H_0(\emptyset) \simeq \mb{Z}$$.

## Defining homology in degree $$-1$$ looks everything but natural.

That is an esthetic argument, purely due to the choice of index. The standard $$n$$-simplex is defined as a subset of $$\mb{R}^{n+1}$$, and so might just as well have been called the $$n+1$$-simplex. In essence, $$n = -1$$ is the true ‘zero-case’ of (co)homology.

## Do we really want to postulate an empty $$-1$$-simplex?

Besides being a building block in the definition of reduced homology, the $$-1$$-simplex is useful in other ways. It allows one to observe that every subset of vertices of a simplex spans a subsimplex (a face), and that, in a simplicial complex, the intersection of any two simplices is a face of both. More generally, the number of $$i$$-faces of an $$n$$-simplex is $$\binom{i+1}{n+1}$$, but the resulting Pascal’s Triangle is not complete without the unique $$-1$$-dimensional face (Figure 2).

\begin{array} {lcccccc} & i=-1 & i=0 & i=1 &i=2 & i=3 & i=4 & \dots\\ n=-1 & 1 & 0 & 0 & 0 & 0 & 0 & \dots\\ n=0 & 1 & 1 & 0 & 0 & 0 & 0 & \dots\\ n\simeq1 & 1 & 2 & 1 & 0 & 0 & 0 & \dots\\ n\simeq2 & 1 & 3 & 3 & 1 & 0 & 0 & \dots\\ n\simeq3 & 1 & 4 & 6 & 4 & 1 & 0 & \dots\\ n\simeq4 & 1 & 5 & 10 & 10 & 5 & 1 & \dots\\ \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \end{array}

Figure 2: The number of $$i$$-faces of an $$n$$-simplex

## So how precisely is traditional homology a special case of reduced homology?

It is possible to give at least two equivalent definitions. First, traditional homology can be subsumed under the more general concept of relative homology (Definition 4).

Definition 4: Let $$X$$ be a topological space, and $$A \sbeq X$$ a subspace. Then for any $$n \in \mb{Z}$$, the relative homology group $$H_n(X,A)$$ is the $$n$$th homology group of the chain complex $$C(X)/C(A)$$.

This definition of relative homology is equally valid for reduced and traditional homology, and this choice does not affect the outcome: $$H_n(X,A)$$ and $$H_n(X,A)_{\emptyset}$$ are isomorphic for all $$n$$. Moreover, both reduced and traditional homology can be derived back from relative homology. For $$x \in X$$ any point, $$H_n(X,\{x\})$$ is the reduced homology group $$H_n(X)$$, while $$H_n(X,\emptyset)$$ is the traditional homology group $$H_n(X)_{\emptyset}$$ (hence this choice of notation). In other words, traditional homology is reduced homology relativised to the empty space.

## One could also define reduced homology as traditional homology relativised to a point.

Except that the choice of a point of $$X$$ makes this arguably less elegant, and it cannot be done for the empty space.

## What is the second equivalent definition?

The traditional homology of a space $$X$$ corresponds to the reduced homology of $$X \sqcup \{z\}$$, where {z} is a singleton.

## That’s not a principled definition, that’s a hack.

More formally, for any $$n \in \mb{Z}$$, we let $${H_n}_\emptyset$$ be the composition $$H_nFI$$, where $$F: \mf{Top}_* \lra \mf{Top}$$ is the forgetful functor and $$I$$ is its left adjoint which sends $$X$$ to $$X \sqcup \{z\}$$. Note that $$I$$ preserves coproducts and sends products to smash products, so this gives us a way to relate traditional and reduced homology.

## If traditional homology is just a derivative of reduced homology, why does it feature in Poincaré Duality? That formula doesn’t work if you plug in reduced homology groups.

But there is a more general Poincaré duality theorem defined in terms of relative homology (Theorem 2, see Theorem 8.3 of Bredon 1993). And since traditional homology is a special case of relative homology, we obtain traditional Poincaré duality when $$L \simeq M$$ and $$K = \emptyset$$.

Theorem 2 (Poincaré–Alexander–Lefschetz): Let $$M$$ be an $$n$$-dimensional oriented manifold and let $$K \sbeq L \sbeq M$$ be compact subspaces of $$M$$. Then for every $$i$$,

$$H_i(M\setminus K, M\setminus L) \simeq H^{n-i}(L,K).$$

## What about the Künneth theorem?

Here too there exists a more general, relative theorem (Theorem 3, see p276 of Hatcher 2002) which reduces to the traditional Künneth theorem when $$A = B = \emptyset$$.

Theorem 3 (Künneth): Let $$X$$ and $$Y$$ be two topological spaces. Then for each $$n$$, there exists a short exact sequence

$$0 \lra \bigoplus_i H_i(X,A) \otimes H_{n-1}(Y,B) \lra H_n(X \times Y,(A \times Y) \cup (B \times X)) \lra \bigoplus_i \Tor(H_i(X,A), H_{n-i-1}(Y,B)) \lra 0$$
which splits.

There also exists a reduced Künneth theorem, which we obtain when $$A = \{x\}$$ and $$B = \{y\}$$ (Corrolary 3.1).

Corrolary 3.1: Let $$X$$ and $$Y$$ be two non-empty topological spaces. Then for each $$n$$, there exists a short exact sequence

$$0 \lra \bigoplus_i H_i(X) \otimes H_{n-1}(Y) \lra H_n(X \wedge Y) \lra \bigoplus_i \Tor(H_i(X), H_{n-i-1}(Y)) \lra 0$$
which splits.

We can also see the traditional Künneth theorem as an implication of this reduced Künneth theorem, given that the traditional homology of $$X$$ and $$Y$$ corresponds to the reduced homology of $$X \sqcup \{z\}$$ and $$Y \sqcup \{z\}$$, and that $$(X \sqcup \{z\}) \wedge_z (Y \sqcup \{z\}) \simeq (X \times Y) \sqcup \{z\}$$.

## The $$i$$th traditional homology group of the $$n$$-fold torus $$T^n$$ is $$\mb{Z}^{\binom{i}{n}}$$. With reduced homology, this pattern breaks down for $$i=0$$. Isn’t this an example where traditional homology produces more natural results?

That the ranks of the homology groups of a torus are binomial coefficients is a direct combinatorial consequence of repeated application of the Künneth theorem to multiplication of $$S^1$$ with itself. One can obtain a similar result with the reduced Künneth Theorem for a whole range of spaces. The reduced equivalent of multiplying $$S^1$$ with itself is taking the smash product of $$S^0 \vee_x S^1 \simeq S^1 \sqcup \{z\}$$ with itself in $$z$$, producing $$S^0 \vee_x T^n \simeq T^n \sqcup \{z\}$$. But one can also take the smash product in other points to obtain different spaces with the same homology. A consistent choice for $$x$$ produces the wedge sum $$\bigvee_{i}\bigvee_{\binom{i}{n}}S^i$$. Alternating choices for $$z$$ and $$x$$ produce various halfway structures, all with homology groups whose ranks correspond to binomial coefficients.

## References

Bredon, Glen Eugene, 1993. Topology and Geometry. Graduate Texts in Mathematics 139. Springer Verlag, New York.
Hatcher, Allen Edward, 2002. Algebraic Topology.