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I'm trying to show that $H_1(\mathbb{R}, \mathbb{Q})$ is free abelian, this is another exercise in Hatcher.

I'm not sure but I thought I can use the exact sequence

$$ \cdots 0 \xrightarrow{f} H_1(\mathbb{R}, \mathbb{Q}) \xrightarrow{g} H_0(\mathbb{Q}) \xrightarrow{h} H_0(\mathbb{R}) \xrightarrow{i} \cdots$$

and then

$$ 0 \xrightarrow{f} H_1(\mathbb{R}, \mathbb{Q}) \xrightarrow{g} H_0(\mathbb{Q}) \xrightarrow{h} \operatorname{ker} (i) \xrightarrow{i}0$$

My idea is to use that an exact sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow$ splits if $C$ is free. So I thought because $\operatorname{ker}(i)$ is a subgroup of $\mathbb{Z}$, it's free. If this is correct, all I need to finish is $H_0(\mathbb{Q})$. I think it should be a countable infinite direct sum of $\mathbb{Z}$'s but I don't know how to express that. How about $\mathbb{Z}^\mathbb{N}$?

Then I'd have $H_0(\mathbb{Q}) = H_1(\mathbb{Q}, \mathbb{R}) \oplus \operatorname{ker}(i)$ and therefore $H_1(\mathbb{Q}, \mathbb{R}) \cong H_0(\mathbb{Q}) / \operatorname{ker}(i) \cong H_0(\mathbb{Q})$.

Now I'm supposed to find a basis for that.

Can someone tell me if my calculations are correct and if yes give me a hint on how to find a basis for $\mathbb{Z}^\mathbb{N}$? Many thanks for your help!

Edit I just found out we have the following exact sequence: $$0 \rightarrow H_1(\mathbb{R}, \mathbb{Q}) \xrightarrow{f} H_0(\mathbb{Q}) \xrightarrow{g} H_0(\mathbb{R}) \xrightarrow{h} 0$$

Using the previous exercise, 16 (a), I know that $H_0(\mathbb{R}, \mathbb{Q}) = 0$ because $\mathbb{Q}$ meets every path-component of $\mathbb{R}$. I also have $H_0(\mathbb{R}) = \mathbb{Z}$.

Edit 2

OK, I think I have it:

$$ H_0( \mathbb{Q} ) = \oplus_{q \in \mathbb{Q}} \mathbb{Z} $$

$$ g: \oplus_{q \in \mathbb{Q}} \mathbb{Z} \rightarrow \mathbb{Z}$$

$$ \operatorname{im}{g} = \mathbb{Z} \implies \operatorname{ker}{g} = \oplus_{q \in \mathbb{Q} - \ast} \mathbb{Z}$$

Here $\mathbb{Q} - \ast$ should be the rationals minus a point.

$$ H_1(\mathbb{R}, \mathbb{Q}) / 0 = H_1(\mathbb{R}, \mathbb{Q}) = \operatorname{im}{f} = \oplus_{q \in \mathbb{Q} - \ast} \mathbb{Z}$$

Can you tell me if this is correct?

What is a basis for this?

Many thanks for your help!

Rudy the Reindeer
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1 Answers1

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A subgroup of a free abelian group is free abelian, by a well known theorem of Kaplanski.

That fact and the long exact sequence do the trick.

Later. $\newcommand\QQ{\mathbb{Q}}\newcommand\RR{\mathbb{R}}\newcommand\ZZ{\mathbb{Z}}$Let's "compute" $H_1(\RR,\QQ)$.

We know that the path-connected components of $\QQ$ are the points. It follows from the general description of $H_0$ that if we denote, for each $q\in\QQ$, $[q]_\QQ$ the homology class in $H_0(\QQ)$ of the point $q$ and $\ZZ[q]_\QQ$ the subgroup it generates in $H_0(\QQ)$, then there is an isomorphism $$H_0(\QQ)\cong\bigoplus_{q\in\QQ}\ZZ[q]_\QQ$$

We also know that $H_0(\RR)$ is free abelian of rank one, generated by the homology class of any point; moreover, we know that if $x$ and $y$ are points in $\RR$ and we denote $[x]_\RR$ and $[y]_\RR$ their homology classes in $H_0(\RR)$, then $[x]_\RR=[y]_\RR$.

Now, the map $g$ is induced by the inclusion $\QQ\to\RR$. In other words, for each $q\in\QQ$ we have $$g([q]_\QQ)=[q]_\RR.$$ In fact, since the set $\{[q]_\QQ:q\in\QQ\}$ is a basis of the abelian group $H_0(\QQ)$, this information completely determines the map $g$.

Next, and as you observed, there is a short exact sequence $$0\to H_1(\RR,\QQ)\xrightarrow fH_0(\QQ)\xrightarrow gH_0(\RR)$$ coming from the long exact sequence inhomology for the pair $(\RR,\QQ)$. We therefore obtain a description of $H_1(\RR,\QQ)$ as an abelian group isomorphic to the kernel of the map $g$.

Can you construct a basis of $\ker g$?

Mariano Suárez-Álvarez
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  • thank you! But what about my computations? Is $H_0(\mathbb{Q}) = \mathbb{Z}^\mathbb{N}$? – Rudy the Reindeer Sep 06 '11 at 06:11
  • @Matt, $H_0$ of a space is the free abelian group on the path components of the space. What are the path components of $\mathbb Q$? – Mariano Suárez-Álvarez Sep 06 '11 at 06:15
  • Well, that's what I'm very insecure about. Is it or is it not, $\mathbb{q_i}$? Every point in $\mathbb{Q}$ is a path-component of it I think. But then I'd have $\mathbb{Z}^\mathbb{N}$ and I'm not sure about that... – Rudy the Reindeer Sep 06 '11 at 06:40
  • Why would you have $\mathbb Z^{\mathbb N}$, the direct product of countably many copies of $\mathbb Z$? (You don't, by the way) – Mariano Suárez-Álvarez Sep 06 '11 at 06:41
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    In any case, you should get rid of any insecurities about what are the path components of $\mathbb Q$ before continuing... Describe all continuous functions $[0,1]\to\mathbb Q$. – Mariano Suárez-Álvarez Sep 06 '11 at 06:42
  • Because every path-component (i.e. point $q_i$) generates a homology group that is isomorphic to $\mathbb{Z}$. Then I do $H_0(\mathbb{Q}) = \oplus H_0(q_i) = \mathbb{Z}^\mathbb{Q} = \mathbb{Z}^\mathbb{N}$. Where is my mistake? – Rudy the Reindeer Sep 06 '11 at 06:44
  • Every continuous function has to be constant, equal to $q_i$? – Rudy the Reindeer Sep 06 '11 at 06:44
  • @Matt, when one writes $\mathbb Z^A$ with $A$ an infinite set, that means the direct product of copies of $\mathbb Z$, one per element in $A$; what you get, on the other hand, is a direct sum. – Mariano Suárez-Álvarez Sep 06 '11 at 06:45
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    Don't ask me: give me a proof! :) – Mariano Suárez-Álvarez Sep 06 '11 at 06:45
  • If it maps to more than one point, say to two, $q_1, q_2$, then for every delta their distance will be greater equals $|q_1 - q_2|$. So pick epsilon smaller than this distance and you cannot find a delta such that continuity holds... – Rudy the Reindeer Sep 06 '11 at 06:50
  • So what's the difference between the direct sum and the direct product in this case? I thought they were the same! – Rudy the Reindeer Sep 06 '11 at 06:52
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    You should try to prove in detail that every continuous function $[0,1]\to\mathbb Q$ is constant---what you wrote is not a correct argument. This is important: homology theory is a complicated elaboration of that very fact! – Mariano Suárez-Álvarez Sep 06 '11 at 06:54
  • The difference between a direct product and a direct sum is stated, for example, here: http://en.wikipedia.org/wiki/Direct_sum_of_modules#Properties – Mariano Suárez-Álvarez Sep 06 '11 at 06:55
  • There are so many words I don't know the definition of! But so $H_0(\mathbb{Q}) = \oplus_{q_i} \mathbb{Z}$? Would that be correct if I wrote it like that? – Rudy the Reindeer Sep 06 '11 at 06:59
  • OK, I have a better argument. Continuous functions map connected spaces to connected spaces. $[0,1]$ is connected, $\mathbb{Q}$ is not, therefore there cannot be a continuous function $[0,1] \rightarrow \mathbb{Q}$? – Rudy the Reindeer Sep 06 '11 at 07:29
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    The difference between product and sum is that sums allow only for strings with finite support, but products allow for infinite support; support is the number of non-zero terms in the string. – gary Sep 06 '11 at 07:51
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    Assume there is , e.g., a continuous function $f$: from $X$ connected into ${0,1}$ discrete. Then each of {$0$}, {$1$} is open in {$0,1$}, so that each of their respective inverse images is open in X. Then $f^{-1} 0 \cup f^{-1} 1 $ is a disconnection of X. But not only is $\mathbb Q$ not connected; it is, as you said totally disconnected , i.e., its components are the singletons. – gary Sep 06 '11 at 07:54
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    Note that I am not saying that the rationals are a discrete space; they are not (e.g., singletons are not open), it was just an idea for the proof. – gary Sep 06 '11 at 10:13
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    Dear Matt, I just want to emphasize Mariano's point above: you should abandon thinking about this problem for the moment, and focus on computing the path components of $\mathbb Q$. This is something that can be proved from first principles, just using the ideas of an undergrad real analysis course. Regards, – Matt E Sep 06 '11 at 18:25
  • So a basis for $\operatorname{ker}{g}$ is ${ [q]_\mathbb{Q} : q \in \mathbb{Q} - \ast }$? – Rudy the Reindeer Sep 07 '11 at 07:22