Why Is $H_*(pt) \cong \mathbb{Z}$ given $H_0(pt) \cong \mathbb{Z}$, $H_n(pt) \cong 0 \ n>0$ ? Reduced homology definition.

Question

In nLab and other MSE posts for questions relating to the isomorphism $H_*(X) \cong \tilde{H_*}(X) \oplus H_*(pt)$ they all have to do with some augmentation map $C_0(X) \to \mathbb{Z}$. However, my professor defined it differently and I am confused on its definition. It goes basically as follows: $pt$ is a terminal object in $Top$, so for a topological space $X \neq 0$ there exists the map $\epsilon: X \to pt$, which is a split epimorphism. This induces the map on homology $\epsilon_*:H_*(X) \to H_*(pt)$ (this is just a chain map between $H_*(X)$ and $H_*(pt)$ right?) and we know $H_*(pt) = \mathbb{Z}$ if $* = 0$ and $0$ else. Since the induced map $\epsilon_*$ must also be a split epimorphism, we have that $H_*(X) \cong \ker(\epsilon_*) \oplus H_*(pt)$ where we defined the reduced homology of $X$, $\tilde{H_*}(X) := \ker(\epsilon_*)$. But then later in the notes it seems he exclusively switches to using $\mathbb{Z}$ for $H_*(pt)$, which I also see in Hatcher and other materials. Is this just abuse of notation in a way? In my understanding $H_*(pt)$ is the chain $\cdots \to 0 \to 0 \to \mathbb{Z} \to 0$ that results from the chain complex $(H_n(pt),\partial$). How then can we write that $H_*(pt) \cong \mathbb{Z}$? Viewing $H_*(pt)$ as just a graded abelian group its not so clear to me how you can direct sum the reduced homology and $\mathbb{Z}$ and that this is well defined. Do I have an mis understanding of what $H_*$ really means?

Answer 1

As Ben Steffan comments, $H_*(Z)$ is a graded (abelian) group. Such an object is nothing else than a collection $\mathbf G = (G_n)_{n \in \mathbb Z}$ of abelian groups $G_n$ indexed by $\mathbb Z$. A graded group does not have differentials $\partial : G_{n+1} \to G_n$ and thus is no chain complex. [Okay, if you want, you can regard a graded group as a chain complex with zero differentials, but this is just a formal trick.]

For each non-empty $X$ we have $$H_*(X) \approx \ker(\epsilon_*) \oplus H_*(pt)$$ where the direct sum of graded groups is taken levelwise. The proof does not use any particular features of singular homology; it is based only on the fact that that the $H_n$ are functors. Thus it works for each generalized homology theory $H_*$. Such a theory is required to satisfy all Eilenberg-Steenrod axioms except the dimension axiom. See for example

Switzer, Robert M. Algebraic topology - homotopy and homology. Springer, 2017.

Therefore $\tilde H_*(X) := \ker(\epsilon_*)$ is a very general definition not depending on a sprecific construction of homology groups (like singular homology groups which are based on the singular chain complex).

For singular homology we have $H_0(pt) = \mathbb Z$ and $H_n(pt) = 0$ for $n \ne 0$. Writing $H_*(pt) = \mathbb Z$ is an abuse of notation. However, one could adopt the convention that each abelian group $G$ is regarded as the graded group $\mathbf G = (G_n)_{n \in \mathbb Z}$ with $G_0 = G$ and $G_n = 0$ for $n \ne 0$. Doing so produces $$H_*(X) \approx \tilde H_*(X) \oplus \mathbb Z$$ for singular homology and non-empty $X$.

It is not a priori clear that the above definition of $\tilde H_*(X)$ agrees with the usual definition of reduced singular homology groups via the augmented chain complex. This is trivial for $n \ne 0$. For $n = 0$ see my answer to Why is $0 \to \tilde{H}_0(X) \to H_0(X) \to H_0(\{x\}) \to 0$ exact?