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This is the dual of Use of Reduced Homology. The answers to that question and the fact that basically every homological argument I have seen so far gets easier when using reduced homology left me with the question: Why do we introduce non-reduced homology in the first place?

One argument seems to be that (say singular) non-reduced homology is slightly easier to define, because you don’t need the augmentation map. However, if I’m not mistaken, you can also get reduced homology by defining the empty set to be a $(-1)$-simplex (and it is a face of every simplex in precisely one way) and then proceeding in the same way as you do for normal homology. This seem hardly more difficult.

Is there another reason to prefer non-reduced homology?

Eike Schulte
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2 Answers2

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Not exactly a use, but a more category theoretic perspective of the difference between reduced and unreduced homology (and hence possibly different applications). In the category of topological spaces (let’s call this Top), we have a coproduct given by disjoint union. However in the category of pointed topological spaces (let’s call this pTop) the coproduct is the wedge sum.

The relevant homology functors preserve coproducts. We see this in the isomorphisms $$H_\ast(\coprod_i X_i) \cong \bigoplus_i H_\ast(X_i)$$$$\tilde{H}_\ast(\bigvee_i X_i) \cong \bigoplus_i \tilde{H}_\ast (X_i)$$

Thus, different homology theories may be necessary depending on the category we are working in, and if we want to use some functorial properties. I'm not sure if you'll count this as a use but it may illustrate why we study both theories (and why reduced isn't just regular homology with a $\mathbb{Z}$ removed in dimension $0$).

Also note that the Kunneth theorem in unreduced homology is a fairly straightforward statement (let's say over a field for simplicity) that $H_\ast(X \times Y) \cong H_\ast(X) \otimes H_\ast(Y)$ but this statement isn't so straightforward for reduced homology. The missing $\mathbb{Z}$ in $0$-dimensional homology shows up in multiple homology groups of the product and so it isn't a straightforward computation in the tensor product.

The key in both cases is what exactly do you want to do with your homology theory?

Alex Provost
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Osama Ghani
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I also have the impression that reduced homology is the correct definition whereas usual homology misses elementary facts about the empty set: As indicated in the question, the point is how to define chains of order $-1$. An $n$-simplex is the convex hull (in some Euclidean space) of $n+1$ points in general position, hence the $-1$-simplex is empty. There is precisely one map $e:\emptyset\to X$ and hence the group $C_{-1}(X)$ of $-1$-chains is the free group with generator $e$. What usually is called the augmentation morphism $\varepsilon$ then just becomes the usual boundary map $\partial_0:C_0(X)\to C_{-1}(X)$.

(In the same spirit, all other chain groups $C_{-n}(X)$ are singly generated, and the boundary maps become the identities.)

In the category of pointed topological spaces the difference is that the empty map $e:\emptyset\to X$ is not a morphism of the category, so that in this case $C_{-1}(X)$ is the abelian group with empty generator, i.e., the trivial group.

Nubtom
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Jochen
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