In Rotman he defined the Augmented singular complex by extending the singular chain complex of a space $…\rightarrow S_2(X) \rightarrow S_1(X) \rightarrow S_0(X) \rightarrow 0$ By defining $ \epsilon (\sum m_x x)= \sum m_x[\emptyset]$.
In spanner he says this map $\epsilon$ must be subjective hence, possibility of considering $X=\emptyset$ goes away.
Hatcher says, we should choose $X$ to be nonempty to avoid getting nonzero homology groups of negative degree.
But Rotman never mentions anything about emptiness of $X$.
Later he gives a problem which says,
If $A\subset X$ , then there is an exact sequence $…\rightarrow \tilde H_n(A)\rightarrow \tilde H_n(X) \rightarrow H_n(X,A)\rightarrow …$ , which ends at
$…\rightarrow \tilde H_0(A) \rightarrow \tilde H_0(X) \rightarrow H_0(X,A) \rightarrow 0$
For, $A\neq \emptyset$ the problem is plain coming from the equality of chain complexes , $\tilde S_*(X)/\tilde S_*(A) = S_*(X)/S_*(A)$
If I put $A=\emptyset$ then depending upon $X$ to be non empty or empty various cases are coming and some of them are contradictory.
For example, if $A=\emptyset$ and $X\neq \emptyset$ then from the exact sequence $\tilde H_0(A)=0, H_0(X,A)= H_0(X)$ so, $\tilde H_0(X)\cong H_0(X)$, but this contradicts the relation of 0th reduced homology and 0th homology group in terms of rank, when X has finitely many path components.
Is there any way to deal with this?
Does the usual practice of reduced homology groups deal with nonempty spaces only?
Best Answer
The augmented chain complex of a space $X$ is $$...\rightarrow S_2(X)\stackrel{\partial}{\rightarrow} S_1(X) \stackrel{\partial}{\rightarrow} S_0(X) \stackrel{\epsilon}{\rightarrow} \mathbb Z \to 0 .$$ This is defined also for $X = \emptyset$, but in fact is usually only considered for non-empty $X$. The reduced homology groups of $X$ are the homology groups of the augmented chain complex, therefore we have $\tilde H_n(X) = H_n(X)$ for $n > 0$. For $n = 0$ we get $\tilde H_0(X) = \ker(\epsilon)/\text{im}(\partial)$ which can be identified with a subgroup of $H_0(X) = S_0(X)/\text{im}(\partial)$. Moreover, one can easily show that $\tilde H_0(X) \approx \ker(p_* : H_0(X) \to H_0(*))$, where $p : X \to *$ is the unique map to a one-point space $*$. Note that $\tilde H_0(X) = 0$ for $X = \emptyset$.
What happens for $n = -1$? If $X \ne \emptyset$, then $\epsilon$ is surjective and $\tilde H_{-1}(X) = 0$, but if $X = \emptyset$, then we get $S_0(X) = 0$ and $\tilde H_{-1}(X) = \mathbb Z$. This is why Hatcher says that we should choose $X$ to be non-empty to avoid getting nonzero reduced homology groups of negative degree. However, it is no real problem to allow also $X = \emptyset$.
Without assuming $A \ne \emptyset$, the exact sequence of $(X,A)$ $$…\rightarrow \tilde H_n(A)\rightarrow \tilde H_n(X) \rightarrow H_n(X,A)\rightarrow …$$ ends with $$…\rightarrow \tilde H_0(A) \rightarrow \tilde H_0(X) \rightarrow H_0(X,A) \rightarrow \tilde H_{-1}(A) \rightarrow \tilde H_{-1}(X) \to 0$$ You see that for $A \ne \emptyset$ we get $\tilde H_{-1}(A) = \tilde H_{-1}(X) = 0$ which yields Rotman's and Hatcher's sequence. For $A = \emptyset$ we get $$…\rightarrow \tilde H_0(A) \rightarrow \tilde H_0(X) \rightarrow H_0(X,A) \rightarrow \mathbb Z \rightarrow \tilde H_{-1}(X) \to 0$$ where $\tilde H_{-1}(X) = 0$ if $X \ne \emptyset$ and $\tilde H_{-1}(X) = \mathbb Z$ if $X = \emptyset$.
You see that we do not get contradictions.