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I'm puzzled by the definition of inverse limit in this Wolfram article.

I thought if an object was defined by a universal property it meant that the object is unique up to unique isomorphism. This would mean that in the article I linked, $\alpha$ should be a unique isomorphism not just a homomorphism as written in the article. Is this a typo?

And I have another question: if someone asks me what an inverse limit is, would I be giving a correct answer if I answered as follows:

The inverse limit of an inverse system $(X_i, f_{ij})$ is the unique pair $(X, \pi_i)$ (where $X$ is an object and $\pi_i$ are morphisms such that $f_{ij} \circ \pi_i = \pi_j$) such that for any pair $(Y, \pi_i^\prime)$ (with $f_{ij} \circ \pi_i^\prime = \pi_j^\prime$) there exists a unique isomorphism $u^{-1}: X \to Y$ such that the following diagram commutes:

enter image description here

Thanks for your help.

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No, $\alpha$ is supposed to be just a morphism. Their definition is correct. –  Zhen Lin Aug 1 '12 at 9:34
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See also the categorial definition of limit in general. –  Martin Sleziak Aug 1 '12 at 9:36
    
@MattN I suppose you follow meta, but I'll write a comment anyway. I've noticed that during the last month you asked (on some days) 3 questions or more, so I wanted to make sure that you know about 50 questions/month (=30 days) quota, which has recently been implemented at MSE, see meta. –  Martin Sleziak Aug 1 '12 at 10:07
    
Dear @MartinSleziak, thank you very much for this. No, I don't follow meta. I am very grateful for pointing this out to me. –  Matt N. Aug 1 '12 at 10:10

1 Answer 1

up vote 3 down vote accepted

You are mixing two things here:

1) In general there is only a unique homomorphism not an isomorphism.

2) The inverse limit is still unique up to unique isomorphism.

The reason that you still have 2) is that if you had inverse limits $X$ and $Y$ then you had the universal property for both of them. So you have unique homomorphisms $X\to Y$ and $Y\to X$. You can show that their compositions are the identities (by invoking the universal property for a third time.)

Edit To clarify regarding the comments: You have the universal propery for an object $X$. Then you can test it for any objects $Y$ which comes with the according $\pi_i$. You get a homomorphism $f:Y\to X$. Note that $Y$ can be a fairly arbitrary object. Only if $Y$ also has the universal property you get a map in the other direction which then turns out to be an inverse.

Edit2: Your example is pretty good so I'll work with it. So let $X_n=[-\frac1n,\frac1n]$ with the obvious inclusions as maps between them. Which candidates for the inverse limit do we have? First note that $X$ comes with maps (i.e. inclusions) to all of the $X_n$, so $X$ is a subset of every $X_n$ and therefore a subset of the intersection. In our example $X$ can only be $\{0\}$ or $\emptyset$, since those are the only subsets of the intersection. Note that the class of all candidates at this point is the class of all "$Y$" at which we have to test in the next step.

Now we turn to the universal property: for any test-set $Y$ which is a subset of the intersection we want to have a morphism (inclusion) $Y\subset X$. Thus $X$ must be contained in the intersection (step 1) and also contain all other sets with this property (step 2), which leaves us with $X=\{0\}$.

This is somewhat the whole idea of what is going on there. In general $X$ has to be "small enough" to come with maps into all the $X_n$, and then it must be the "biggest" among all objects with this property.

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But isn't the last part in your answer exactly the same as saying they are two sided inverses which means it's an isomorphism? –  Matt N. Aug 1 '12 at 9:50
    
I mean I could replace "homo" by "iso" in the universal property, no? –  Matt N. Aug 1 '12 at 9:55
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I answered you comments in the edit. I hope this clarifies things. If you are unsure you can always try an easy example. Maybe the category of sets with inclusions as morphisms. Then the inverse limit $X$ will be the intersection of the $X_i$, but you still get unique morphisms for any $Y\subset X$ to $X$. –  Simon Markett Aug 1 '12 at 10:06
    
Dear @Simon, thank you very much for this example. I tried to think about $X_n = [-\frac1n , \frac1n]$ with $i_n : X_n \hookrightarrow X_{n-1}$. How do I compute the inverse limit of this? I don't really see how it should be the intersection of $X_n$. –  Matt N. Aug 1 '12 at 15:23
    
Ok, I think I understand both, what I mixed up and also how to see that it's the intersection. –  Matt N. Aug 1 '12 at 18:27

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