$\mathbb Z\oplus\mathbb Z\cong H_1(U_1)\oplus \mathbb Z$ then why $H_1(U_1)\cong \mathbb Z$

Question

https://math.stackexchange.com/a/1171748/1015445

Here Pedro A. Castillejo uses $$\dots \to H_{n+1}(\Bbb R^3) \to H_n(T^2) \to H_n(U_1) \oplus H_n(S^1) \to {H}_{n}(\Bbb R^3 ) \to H_{n-1}(T^2) \to \dots$$ Then

$$H_2(U_1) \simeq H_1(U_1) \simeq \Bbb Z$$ is also okay because $$H_n (T^2) \simeq H_n(U_1) \oplus H_n(S^1)$$

Which implies $$\mathbb Z^2\cong H_1(U_1)\oplus \mathbb Z$$

However how we can conclude (cancels $\mathbb Z$) and say $$H_1(U_1)\cong \mathbb Z$$

because cancelation is not correct in general and I am not satisfied.

Answer 1

Sure. It follows from the classification of finitely generated abelian groups for example. See: https://en.wikipedia.org/wiki/Finitely_generated_abelian_group.

Firstly, $H_1(U_1)$ is a finitely generated abelian group. Finitely generated follows since $U_1$ is a topological space of finite topological type and homology groups are always abelian. The equation implies it is rank one since rank of f.g. abelian groups is additive over direct products.

Since it is isomorphic to a subgroup of the torsion free group $\mathbb{Z}^2$, it is torsion free. Hence it is $\mathbb{Z}$ by the classification of finitely generated abelian groups.

Answer 2

Nick's answer is fine, but here's an answer using very little machinery.

We begin with a lemma:

Lemma: Suppose $(a,b)\in \mathbb{Z}^2$ with $a$ and $b$ relatively prime. There there is a group isomorphism $\phi_{(a,b)}:\mathbb{Z}^2\rightarrow \mathbb{Z}^2$ with $\phi_{(a,b)}(a,b) = (0,1)$.

Proof: Because $a$ and $b$ are relatively prime, there are integers $s$ and $t$ for which $as + bt = 1$. Writing elements of $\mathbb{Z}^2$ as column vectors, we define $\phi_{(a,b)}$ to be multiplication by $\begin{bmatrix} b & -a \\ s & t\end{bmatrix}$. The fact that $\phi_{(a,b)}$ is an isomorphism mapping $\begin{bmatrix} a\\b\end{bmatrix}$ to $\begin{bmatrix} 0\\ 1\end{bmatrix}$ is left to the reader. $\square$

Now, fix an isomorphism $f:H_1(U_1)\oplus \mathbb{Z}\rightarrow \mathbb{Z}^2$. By replacing $f$ by $\phi_{f(0,1)}\circ f$ if necessary, we may assume without loss of generality that $f(0,z) = (0,z)$ for all $z\in \mathbb{Z}$.

Let $\pi:\mathbb{Z}^2\rightarrow \mathbb{Z}$ with $\pi(c,d) = c$. I claim that $g:=(\pi\circ f)|_{H_1(U_1)\oplus \{0\}}$ is an isomorphism between $H_1(U_1)\oplus \{0\}$ and $\mathbb{Z}$. Since obviously $H_1(U_1)\oplus \{0\}\cong H_1(U_1)$, this gives the desired isomorphism.

I will only sketch the proof that $g$ is a bijection. To see $g$ injective, note that if $(x,0)$ is in the kernel of $g$, then $f(x,0) =(0,d) = f(0,d)$ for some $d\in \mathbb{Z}$. Then use the fact that $f$ is injective.

To see $g$ is surjective, first compute that $\pi(f(x,z))$ is independent of $z$ and conclude that $g$ is surjective if and only if $\pi \circ f$ is. But both $\pi$ and $f$ are surjective, so $\pi \circ f$ obviously is as well.