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I am trying to solve this exercise and wanted to check my approach. Question: Show $H_{1}(\mathbb{R}, \mathbb{Q})$ is free abelian and find a basis

We get the LES from pairs $H_{1}(\mathbb{R}) \longrightarrow H_{1}(\mathbb{R}, \mathbb{Q}) \longrightarrow H_{0}(\mathbb{Q}) \longrightarrow H_{0}(\mathbb{R})$.

Now the reals are path-connected so $H_{1}(\mathbb{R}) = \mathbb{Z}$ which is generated by any point, hence the map $H_{0}(\mathbb{Q}) \longrightarrow H_{0}(\mathbb{R})$ is subjective. We also get that the map $H_{1}(\mathbb{R}, \mathbb{Q}) \longrightarrow H_{0}(\mathbb{Q})$ is injective from $H_{1}(\mathbb{R}) = 0$. Hence using the $1^{st}$ isomorphism theorem, exacteness, and the fact that the domain of an injective map is isomorphic to its image, we get $\mathbb{Z} \cong \frac{H_{0}(\mathbb{Q})}{H_{1}(\mathbb{R}, \mathbb{Q})}$.

Any path joining two rationals must intersect an irrational by the IVT and the density of $\mathbb{R} \setminus \mathbb{Q}$ in the reals. Hence I believe that $H_{0}(\mathbb{Q})$ is a free module with an infinite basis, as the path-components of $\mathbb{Q}$ are simply the rationals themselves and the $0$-th homology is the direct sum of $\mathbb{Z}$, one for each path-component. Hence $H_{1}(\mathbb{R}, \mathbb{Q})$ is isomorphic to a subgroup of a free $\mathbb{Z}$-module where $\mathbb{Z}$ is a PID, thus we deduce that $H_{1}(\mathbb{R}, \mathbb{Q})$ is itself free.

From $\mathbb{Z} \cong \frac{H_{0}(\mathbb{Q})}{H_{1}(\mathbb{R}, \mathbb{Q})}$, we know that $H_{0}(\mathbb{Q}) \cong \mathbb{Z} \bigoplus H_{1}(\mathbb{R}, \mathbb{Q})$.

I am left with finding a basis for $H_{1}(\mathbb{R}, \mathbb{Q})$, which I know must be infinite. Any hint?

thorwi
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1 Answers1

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In the reduced variant, we get that $\widetilde{H}_0(\mathbb{R})=0$, and so the boundary map $\partial:\widetilde{H}_1(\mathbb{R},\mathbb{Q})\to\widetilde{H}_0(\mathbb{Q})$ is an isomorphism. This map (c.f. [Hatcher] p.117) sends the homology class $[\alpha]$ of relative cycle $\alpha$ (i.e., $\alpha=[a,b]\in C_1(\mathbb{R})$ and $\partial_n(\alpha)=b-a\in C_0(\mathbb{Q}), \text{ or }, a,b\in \mathbb{Q}$) to the homology class $[b-a]$ of $\partial_n(\alpha)(=b-a)$ in $\widetilde{H}_0(\mathbb{Q})$ (note that "$-$" doesn't mean standard rational subtraction, but the formal chain subtraction). By fixing $b\in\mathbb{Q}$ we conclude that $\{[b-a]\}_{a\in\mathbb{Q}\backslash\{b\}}$ is a set of free generators of $\tilde{H}_0(\mathbb{Q})$ (note that $a=b$ is ignored because $[b-b]=0$, but in the reduced version we exclude one fixed path component from $H_0(\mathbb{Q})$). Therefore $\tilde{H}_1(\mathbb{R},\mathbb{Q})$ is freely generated by simplices of the form $\{[a,b]\}_{a\in\mathbb{Q}\backslash\{b\}}$.

This translates to non-reduced version since $\tilde{H}_*(X,A)=H_*(X,A)$. And in the non-reduced version the missing "$a=b$" corresponds to the $\mathbb{Z}$ direct summand of $H_0(\mathbb{Q})$.

Anthony
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freakish
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  • I am not sure to understand why the cocycle $H_{1}(\mathbb{R}, \mathbb{Q})$ is an interval with endpoints in $\mathbb{Q}$, we are working with real intervals modulo rational intervals as potential co-cycles (here an interval is a 1-simplex). Could you expand on that? I thought the 1-chains remaining in the quotient $\frac{C_{1}(\mathbb{R})}{C_{1}(\mathbb{Q})}$ would be intervals with endpoints in $\mathbb{R} \setminus \mathbb{Q} $. the boundary map takes the interval to map it a 0-simplex represented by $b-a$, which is a map $\text{point} \rightarrow \mathbb{Q}$ evaluating at point $b-a$. – thorwi Jan 18 '22 at 11:15
  • @thorwi if you look at the definition of homology, you get chain complex $C_n(X,A)\to C_{n-1}(X,A)$. A relative cycle (btw not co-cycle) is then an element from kernel of that map, also called boundary map. Which means the boundary of relative cycle is $0$. And this means that the boundary belongs to $A$ (up to relative homotopy anyway). When chain has a single element, a simplex, this means that simplex has boundary fully in $A$. – freakish Jan 18 '22 at 11:36
  • @thorwi Also the boundary map takes $[a,b]$ interval to $b-a$ chain. This is not a map that evaluates to $b-a$ point. Chains are not maps. Chains are formal sums of maps. And so $b-a$ is $b-a$, it does not evaluate to single point, the concept of evaluation doesn't really work here. It catches the idea that boundary of $[a,b]$ is ${b,a}$ with a specific orientation. – freakish Jan 18 '22 at 11:37
  • It makes sense, thanks! – thorwi Jan 18 '22 at 11:47