2
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ That's a good question, this convention can be tricky. You should try to see what happens if you always put a $\mathbb Z$ as $S_{-1}(X)$, even if $X$ is empty (thus allowing one negative homology group for emptyspaces). $\endgroup$ May 9, 2020 at 8:05

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .