Homology question and reduced homology

Question

I'm looking at rotman, page 103. May someone elaborate why the following holds:

Since $H_0(X)\cong \tilde{H}_0(X)\oplus \mathbb{Z}$ then $H_0(X)\cong \tilde{H}_0(X,x_0)$..

Does the following more general result hold:

If $F$ is free abelian and $F\cong F'\oplus \mathbb{Z}$, where $F'$ is abelian then $F\cong F'$?

May someone elaborate on why either holds?

Also, it's hard to answer the homology part of this without having the context. — Randall, Jun 26 '21 at 19:01
@Randall The result i'm trying to prove is $H_0(X)\cong \tilde{H_0(X)}$ from $H_0(X)$ is isomorphic to direct sum... — , Jun 26 '21 at 19:02
But that's false. Do you mean to compare homology and reduced homology in degrees above $0$? — Randall, Jun 26 '21 at 19:04
Note that $\mathbb Z \oplus \mathbb Z$ is not isomorphic to $\mathbb Z$. Perhaps you meant $H_0(X) \cong \tilde{H}0(X+)$, where $X_+=X \coprod *$? — Andres Mejia, Jun 26 '21 at 19:14

Paul Frost · Answer 1 · 2022-11-05T00:07:51.600

Does the following more general result hold:

If $F$ is free abelian and $F\cong F'\oplus \mathbb{Z}$, where $F'$ is abelian, then $F\cong F'$?

No. Take $F = \mathbb Z \oplus \mathbb Z$ and $F' = \mathbb Z$. I am sure you mean

If $F$ is free abelian and $F \oplus \mathbb Z \cong F'\oplus \mathbb{Z}$, where $F'$ is abelian, then $F\cong F'$?

Here the answer is yes. $G= F\oplus \mathbb Z$ is free abelian and $F'$ embeds as a subgroup into $G$, thus $F'$ is free abelian. If $F$ has finite rank $n$, then $G$ has rank $n+1$, thus $F'\oplus \mathbb{Z}$ has rank $n+1$ and we conclude that $F'$ must also have rank $n$. Similarly, if $F$ has infinite rank, then $G$ has the same rank rank as $F$ and so has $F'\oplus \mathbb{Z}$. We conclude that $F'$ has the same rank as $F$.

Now look at the exact sequence $$\ldots \to H_0(x_0) \to H_0(X) \to H_0(X,x_0) \to 0 .$$ This gives a split exact sequence $$0 \to H_0(x_0) \to H_0(X) \to H_0(X,x_0) \to 0 $$ since the unique map $c : X \to \{x_0\}$ induces a right inverse for $H_0(x_0) \to H_0(X)$. We conclude $$\tilde{H}_0(X)\oplus \mathbb{Z} \cong H_0(X) \cong H_0(X,x_0) \oplus H_0(x_0) \cong H_0(X,x_0) \oplus \mathbb Z$$ which implies $$\tilde{H}_0(X) \cong H_0(X,x_0) .$$

score 0 · Answer 2 · answered Jun 27 '21 at 13:28

Note that $x_0$ is a retract of $X$. Hence $H_0(X)\cong H_0(X,x_0)\oplus H_0(x_0)$ . Since singletons are path connected, $H_0(x_0)\cong \mathbb{Z}$, and so we have $H_0(X,x_0)\oplus H_0(x_0)\cong H_0(X,x_0)\oplus \mathbb{Z}$. Now, as you've shown, we have $H_0(X)\cong \tilde{H}_0(X)\oplus \mathbb{Z}$. In particular we have the following isomorphisms:

$H_0(X)\cong H_0(X,x_0)\oplus \mathbb{Z}\cong \tilde{H}_0(X)\oplus \mathbb{Z}$ .

Since $H_0(X,x_0)$ is free abelian, the result follows.

Homology question and reduced homology

2 Answers2

Linked