# Use of Non-Reduced Homology

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?

• With field coefficients, non-reduced homology is a coalgebra. – John Palmieri Jun 15 '18 at 14:42

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.