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Let $G$ be a topological abelian group and let $\widehat{G}$ denote its completion (i.e. equivalence classes of Cauchy sequences). Let $G_n$ be a descending sequence of subgroups, i.e. $G = G_0 \supset G_1 \supset \dots$ such that $G_n$ form a countable neighborhood basis of $0$.

Let $\bar{x} \in \widehat{G}$ and let $x$ be its representative so that $x$ is Cauchy in $G$. We know that for a given $k$, $x_n$ is constant in $G/G_k$ for $n$ large enough, say $x_n = \xi_k$ in $G/G_k$. So from the sequence $x$ we get a new sequence $\xi$.

Define $\varphi_n : \widehat{G} \to G/G_n$ as $x \mapsto \xi_n$. Then define $\varphi: \widehat{G} \to \prod_n G/ G_n$ as $x \mapsto \xi$. Now we want to show that $\mathrm{Im}\varphi \subset \displaystyle \lim_{\longleftarrow} G/G_n$, i.e. $\theta_i \circ \varphi_i = \varphi_{i-1} $.

For this note that if $x_n = \xi_k$ in $G/G_k$ and $x_n = \xi_{k-1}$ in $G/G_{k-1}$ then for the projections $\theta_{i}: G/G_i \to G/G_{i-i}$ we get $\theta_{i} \varphi_{i}(x) =\theta_{i} \xi_{i} = \xi_{i-1} = \varphi_{i-1}(x)$.

Now to show that $\varphi$ is an isomorphism we want to find a two-sided inverse. I was thinking that if $(\xi_n) \in \displaystyle \lim_{\longleftarrow} G/G_n$ then I could just define $\psi: (\xi_n) \mapsto \overline{(\xi_n)}$ where $\overline{(\xi_n)}$ denotes the equivalence class of $(\xi_n)$ in $\widehat{G}$ but this all feels a bit shaky.

What is the correct definition of inverse of $\varphi$? Thanks for your help.

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    Did try to verify that $\widehat{G}$ satisfies the universal property of $\varprojlim G/G_n$? It looks to me as if you had almost everything in hands to get that. PS: you can get the $\varprojlim$ symbol by using `\varprojlim`2012-07-26
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    How can we know whether $\,x_n\,$ is constant in $\,G/G_k\,$ if you haven't yet told us what is $\,G_k\,$ (normal subgroup of finite index, say?)2012-07-26
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    @t.b. No I didn't: I'm on page 104 of Atiyah-Macdonald and the universal property of inverse limits has not been mentioned so far. And looking at the other two pages of the chapter tells me that it's not in the book at all. I will look it up on Wikipedia. Thanks for the tip with latex.2012-07-26
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    It's a bit confusing: in the book the define an inverse system as a sequence of groups $A_n$ together with a bunch of homomorphisms $\theta_{n, n-1} : A_n \to A_{n-1}$ whereas on [Wikipedia](http://en.wikipedia.org/wiki/Inverse_limit#Algebraic_objects) they additionally require that $\theta_{n, n}$ be the identity and $\theta_{k, m} \circ \theta_{n, k} = \theta_{n, m}$.2012-07-26
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    The is no reason for there to be an isomorphism in the situation you describe, unless the sequence of subgroups is picked with care...2012-07-26
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    Re: "bit confusing," there is no fundamental difference between definitions here. Given the morphisms $\theta_{n,n-1}$ we may construct all of the other morphisms via composition (and placing identity morphisms on each object). However, they are no longer "the same" when the poset indexing the objects of the inverse system is no longer locally finite like $\Bbb N$ (indeed only one definition makes sense then).2012-07-26
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    @anon I don't see how to get $\theta_{n,n} = id_{A_n}$. $\theta_{k,i}$ are just group homos, they could be anything. And they also only go "downwards" in the book: we don't have $\theta_{n-1, n}$. How do you get $\theta_{n,n} = id_{A_n}$?2012-07-28
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    You "get" $\theta_{n,n}=\mathrm{id}_{A_n}$ by simply defining it that way a priori in the definition of an inverse system: you don't need to know any other morphisms to know this follows by definition. Of course $\theta_{n-1,n}$ doesn't designate anything - bringing that up seems off-topic. Finally, given any $n,m\in\Bbb N$ with $n>m$, we can get $\theta_{n,m}$ via iterated composition: $$\theta_{n,m}=\theta_{m+1,m}\circ\theta_{m+2,m+1}\circ\cdots\theta_{n-1,n-2} \circ \theta_{n,n-1}.$$ (I know it seems backwards; put it into a commutative diagram and it will make more sense.)2012-07-28
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    Except I should note that convention varies wildly as to whether the smaller index goes first or second in the subscript of the morphisms, or even subscript vs. supscript. Anyway, this shows that in $\Bbb N,$ to know all morphisms of an inverse system it suffices to know the objects and just the one-off morphisms $\theta_{n,n-1}$. The same idea applies to any locally finite indexing poset (the same "one-off morphisms," but no longer labelled with naturals).2012-07-28
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    @anon Yes, but then you'd have to define $\theta_{n,n}=\mathrm{id}_{A_n}$. But Atiyah-Macdonald don't define $\theta_{n,n}$ at all, they only have $\theta_{n,n-1}$ in their definition of inverse system. Sorry that I still don't understand.2012-07-29
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    I'm telling you the AT definition is equivalent to other definition, because given only the information of the AT definition one can deduce all of the other information needed to fulfill the other definition.2012-07-29
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    @anon I know that you are telling me this! : ) And I've been trying to tell you that I don't believe that we can get a map $\theta_{n,n} = id_{G_n}$ from the AT definition.2012-07-29
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    Not only can you get $\theta_{n,n}=\mathrm{id}_{A_n}$ from the information of the AT definition, you can get it *without any information at all* because it *has* to be that way to satisfy the other definition. There is nothing else it can be.2012-07-29
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    @t.b. I did it with universal properties now. What do you think? : )2012-08-06

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I think you're close to the right map, but let's be careful. If $(x_n) \in \varprojlim G/G_n$ then define a sequence $(y_n)$ of elements in $G$ by taking $y_n$ to be any lift of $x_n$, and then take the equivalence class of this inside $\hat G$. This class does not depend on the choices of lifts, for if $(z_n)$ is another such sequence then $(y_n) - (z_n)$ is a null sequence since $y_n - z_n \in G_N$ for all $n \geq N$.

Let me know how proving that these are inverse goes.

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    Dear Dylan, thank you for this answer! But I don't know what you mean by a lift. Could you explain it to me?2012-07-29
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    @ClarkKent: '$y_n\in G$ is a lift of $x_n\in G/G_n$' just means that the quotient map $G\to G/G_n$ maps $y_n$ to $x_n$.2012-07-29
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    @wildildildlife Yes, thanks, I think now I understand the maps.2012-07-29
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    @Dylan Thanks, I think I understand the maps. For now I'll accept your answer and once I see that you're back on the site I'll have another go at showing that they're inverses.2012-07-29
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    @Clark Will try to have a look later today. Sorry for the silence on my end.2012-08-25
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You could show that your map $\phi$ (which is easily seen to be a group morphism) is injective and surjective.

Injective: if $\phi(x)=0$ in $\varprojlim G/G_n$, then for all $n$, $\phi_n(x)=0$ in $G/G_n$, so for all $n$, eventually $x_i\in G_n$, meaning $(x_i)=0$ in $\hat{G}$.

Surjective: given $\xi=(\xi_n)\in \varprojlim G/G_n$, take lifts $x_n\in G$ of $\xi_n\in G/G_n$; then note that $(x_n)$ is Cauchy in $G$, so yields an element in $\hat{G}$ which by construction gets mapped to $\xi$ by $\phi$.

You probably meant to prove $\hat{G}=\varprojlim G/G_n$ as topological abelian groups. So to finish, you could then prove that $\phi$ is an open map. Alternatively you can just prove that the inverse (which is actually constructed in the proof of surjectivity) is continuous.

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    Ooh, I didn't think of that! Thank you! I thought of the two-sided inverse first because that's what usually works best for stuff involving isomorphisms of tensor products.2012-07-29
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We don't need to construct an isomorphism between the two. We can instead show that $\hat{G}$ satisfies the universal property of $\varprojlim G/G_n$ as follows:

Let our inverse system be $(G/G_n, \theta_{nk})$ (where $n \geq k$), satisfying $\pi_k = \theta_{nk} \circ \pi_n$ where $\pi_n : \hat{G} \to G/G_n$ is the map taking a representative of a Cauchy sequence in $\hat{G}$ and "modding" it by $G_n$ so that all its terms above index $n$ are zero. Clearly, $\pi_n$ are surjective. (We will use this vital property later to construct our unique homomorphism that makes our diagram commute.)

Now let $H$ be any (topological?) Abelian group and let $\pi_n^\prime$ be maps such that $\pi_n^\prime = \theta_{mn} \circ \pi_m^\prime$ for all $m \geq n$. Then we construct a homomorphism $\alpha : H \to \hat{G}$ as follows: Let $h \in H$. Then $\pi_k^\prime (h) \in G/G_k$. Since $\pi_k$ are surjective there exists $g \in \hat{G}$ such that $\pi_k(g) = \pi_k^\prime (h)$. Define $\alpha (h) = g$.

To conclude the proof we need to verify that $\alpha$

(i) is well-defined: let $g,g^\prime$ both be in $\hat{G}$ and such that $\pi_k(g) = \pi_k(g^\prime) = \pi^\prime_k (h)$. We claim that then $g-g^\prime \to 0 $ in $\hat{G}$. Since $g$ and $^\prime$ are equal "mod $G_k$" they agree on the first $k$ terms. By induction they agree on all terms and hence $g-g^\prime$ is the zero sequence.

(ii) is a group homomorphism i.e. $\alpha (h + h^\prime) = \alpha (h) + \alpha (h^\prime)$:

$\alpha (h) = g$ where $g \in \hat{G}$ is such that $\pi_k (g) = \pi_k^\prime(h)$ and $\alpha (h^\prime) = g^\prime$ where $\pi_k (g^\prime) = \pi_k^\prime (h^\prime)$ hence $\alpha (h + h^\prime) = g + g^\prime$ where $g + g^\prime$ is such that $\pi_k(g + g^\prime) = \pi_k (g) + \pi_k (g^\prime) = \pi_k^\prime (h) + \pi_k^\prime (h^\prime) $ and hence $\alpha (h) + \alpha (h^\prime) = g + g^\prime = \alpha(h + h^\prime)$.

(iii) makes the diagram commute: by definition.

(iv) is unique: let $\alpha^\prime$ be such that $\pi_n \circ \alpha^\prime = \pi^\prime_n$ for all $n$. Then $\pi_n \circ \alpha^\prime = \pi^\prime_n = \pi_n \circ \alpha $ which (by same reasoning as in (i)) means that $\alpha$ and $\alpha^\prime$ agree on the first $n$ terms and then by induction they agree on all terms.

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    well, yes, that's a start, but where are the verifications? :)2012-08-06
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    @t.b. Thanks for looking! Ok, ok. But I can't do that right now. I'll do it tomorrow, ok? If I have time. I've got some other stuff I want to do tomorrow as well.2012-08-06
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    @t.b. Done. : )2012-08-07
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    Yes, this looks right, if a bit sketchy. I do not agree with your argument for (ii), however. There's nothing you concatenate here. Verify it directly. Also, I think it would maybe make the argument a bit more transparent if you fusioned (i) and (iv). The description and well-definedness of $\alpha$ also shows its uniqueness.2012-08-07
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    @t.b. Thank you very much for checking it. I added a direct verification to the post.2012-08-07
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    I think I understand what I'm doing, i.e., I understand inverse limits.2012-08-07
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    Yes, this looks okay now. There's some confusion of notation e.g. in point (i) ($g-g' \to 0$ versus $g-g'$ is the zero sequence), but overall you seem to have gotten the main points. Since you announced that you've got other stuff to do I don't insist that you make this all formally correct. I hope you see that there's some room for formal improvement.2012-08-07
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    @t.b. Thank you! Yes, I know there is room for improvement and I will probably do it more formally in the near future. Now I have just posted an attempt at showing direct limits commute with tensor products. If you have time I'd be very grateful if you could have a look at [this](http://math.stackexchange.com/a/179837/26736), because I don't know whether it's right and if it is, I don't know how to finish it.2012-08-07