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I've been reading Hatcher's Algebraic Topology, specifically the paragraph about reduced homology $\tilde{H}_*$ (for singular homology of topological spaces). Can someone please provide reasons why reduced homology is defined and studied?

I understand the following facts, which are all found in Hatcher's book :

$-$ The reduced homology of a point is $0$.

$-$ The reduced homology is the same in all degrees $*$ as the usual singular homology for pairs of spaces $(X,A)$ with $A\neq \emptyset$ : $\tilde{H}_*(X,A)= H_*(X,A)$, and in positive degrees $(*=n>0)$ for single spaces $X$ (that is when $A=\emptyset$). There is the same long exact sequence in reduced homology for a pair of spaces as in standard homology.

$-$ In degree $0$, one has $\tilde{H}_0(X)\oplus\mathbb{Z}\cong H_0(X)$, with the coefficients for homology in $\mathbb{Z}$.

$-$ For any space $X$, and any point $\mathrm{pt}\in X$, there is an isomorphism $\tilde{H}_*(X)\cong H_*(X,\lbrace \mathrm{pt}\rbrace)$

$-$ This in turn implies that when $A\subset U\subset X$ is such that $A$ is closed, $U$ is open, and $A$ is a strong deformation retract of $U$, then there is an exact sequence in reduced homology (that stems from an exact sequence for standard singular homology)

$$\cdots\rightarrow\tilde{H}_*(A)\rightarrow\tilde{H}_*(X)\rightarrow\tilde{H}_*(X/A)\rightarrow\cdots$$

All of this is straightforward to prove, but it doesn't tell me why reduced homology is defined and when it is used. Can someone please shed some light on this matter?

Panini
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Olivier Bégassat
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4 Answers4

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The essential reason for preferring reduced homology (as experts do) is that the suspension axiom holds in all degrees, as it must when one generalizes from spaces to spectra and studies generalized homology theories. Also, when using reduced homology, one need not explicitly use pairs of spaces since $H_*(X,A)$ is the reduced homology of the cofiber $Ci$ of the inclusion $i\colon A\to X$.
The Eilenberg-Steenrod axioms for homology theories have a variant version for reduced theories, and the reduced and unreduced theories determine each other. (See for example my book ``A concise course in algebraic topology'').

Peter May
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Reduced homology is used, mostly, to simplify statements.

For example, it is not true that the homology of a wedge of two spaces $X\vee Y$ is the direct sum of the homologies of $X$ and of $Y$, but the only problem is actually in degree $0$. It is true, on the other hand, that the reduced homology of $X\vee Y$ is the direct sum of the reduced homologies of $X$ and of $Y$. This happens in various other contexts.

N.B.: You should be careful with those isomorphisms you mention, for they are generally not natural.

Mariano Suárez-Álvarez
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    Reduced homology is really the relative homology of $X$ relative to a point, right? So I guess it's natural for it to be related to wedge sum. – Qiaochu Yuan Jul 13 '11 at 02:03
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    The difference lies in that reduced homology is independent of the choice of a point, though :) – Mariano Suárez-Álvarez Jul 13 '11 at 02:08
  • An other good example would be alexander duality. See http://en.wikipedia.org/wiki/Alexander_duality – Alexander Thumm Jul 13 '11 at 09:04
  • could you point out the non canonical isomorphisms please? I can see that the long exact sequence in my last point actually depends on the open neighborhood $U$, so might not be natural. The other ones I think are natural... Am I right? – Olivier Bégassat Jul 13 '11 at 19:08
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    @Olivier: The isomorphism $\tilde H_0(X)\oplus\mathbb Z\cong H_0(X)$ you mention comes from an exact sequence $$0\to \tilde H_0(X)\to H_0(X) \to\mathbb Z\to 0$$ This sequence splits (because $\mathbb Z$ is a projective abelian group, say) (and this splitting gives your isomorphism) but the splitting is not canonical. A splitting corresponds, more or less, to picking an element in $H_0(X)$ which is part of a basis, and this cannot be done, in general, in a natural way. – Mariano Suárez-Álvarez Jul 13 '11 at 20:32
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    Likewise, The isomorphism $\tilde{H}*(X)\approx H*(X,\lbrace \mathrm{pt}\rbrace)$ obviousl depends on the choice of the point $\mathrm{pt}$. – Mariano Suárez-Álvarez Jul 13 '11 at 20:34
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    The isomorphism $\bar{H}0(X)\oplus \mathbb{Z} \cong H_0(X)$ is indeed unnatural, but somewhat counterintuitively (at least for me) is that the isomorphism $\bar{H}_0(X) \cong H_0(X,*)$ is natural as functors $Top* \rightarrow Ab$. And similarly as functors $Top_* \rightarrow Ab$, $\bar{H}_0(X)\oplus \mathbb{Z} \cong H_0(X)$, naturally. So we have an example of two objectwise isomorphic functors that are not naturally isomorphic, that become naturally isomorphic when moving to a slice category. – Connor Malin Aug 07 '20 at 01:59
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Firstly $H_0$ isn't particularly interesting since we frequently deal with connected spaces anyway. Apart from that, in various exact sequences, such as Mayer-Vietoris, using standard homology leaves us with a bunch of $\mathbb{Z}$'s at the end. But exact sequences are so much nicer with $0$'s instead! Reduced homology gives us 'exact'ly this!

Aru Ray
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Reduced homology is the correct notion of homology on pointed spaces.

To explain this, let me start by showing how this works with discrete sets rather than spaces.

Let $N$ be the set $\{1,\dots,n\}$. There's a function $e:N\to\mathbb{R}^n$ that sends each $i$ to the basis vector $e_i$. Now suppose $V$ is any other real vector space, and we have some function $f:N\to V$ that sends the elements of $N$ to arbitrary vectors in $V$. Then there's always a linear map $\hat{f}:\mathbb{R}^n\to V$ that extends $f$ to all of $\mathbb{R}^n$. In other words $\hat{f}$ has the property that $\hat{f}\circ e=f$. Furthermore, $\hat{f}$ is the unique linear map with this property.

The above property is what we mean when we say that $\mathbb{R}^n$ is the free vector space on a set with $n$ elements.

Now, define a pointed set to be a set one of whose elements we have specified as 'the point'. A homomorphism of pointed sets is a function between the sets that sends the point of the source to the point of the target. Note that any vector space is a pointed set in a natural way: we pick the $0$ vector as the point.

Now suppose $N^*$ is a set with $n$ elements, one of which we've chosen as the point. There's a homomorphism $N^*\to\mathbb{R}^{n-1}$ that sends the point to the $0$ vector and sends each other element to a basis vector. Now if $f:N^*\to V$ is any other homomorphism sending the point to $0$ then there's a unique linear map $\hat{f}:\mathbb{R}^{n-1}\to V$ such that $\hat{f}\circ e=f$.

In other words $\mathbb{R}^{n-1}$ is the free vector space on a pointed set with $n$ elements.

You can see that the dimension of the free vector space was reduced by $1$ when we considered pointed sets.

This makes it clear what the reduced homology is. Reduced homology is the correct version of homology to use when our space is naturally equipped with a special point. The reduction in dimension comes from sending that point to $0$.

This is a useful perspective, because it helps us remember when the formulas will be simpler with the reduced or with the normal homology.

  • The quotient $X/A$ naturally has a point: the point that $A$ is quotiented to. So we should use the reduced homology to state the long exact sequence associated with quotients. Likewise for relative homology (which is really just a way to cope with badly behaved quotients).

  • The wedge sum inherently requires the spaces to be pointed, so we have a nice formula in reduced homology, $\tilde{H}(X\vee Y)=\tilde{H}(X)\oplus\tilde{H}(Y)$.

  • On the other hand, the disjoint union of two spaces is not naturally pointed. So there we have a nice formula using the normal homology, $H(X\sqcup Y)=H(X)\oplus H(Y)$.

(Using $\infty$-categories we could make the above precise. The reduced homology is the homology of the free chain complex on a pointed space. But we can appreciate the above without this more abstract perspective.)

Oscar Cunningham
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