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Stuck again. Not even a page through :-(

Reading Awodey's Category Theory [p.17] he says (this is after the definition of what a slice category $\boldsymbol{C}/C$ is):

If $C=P$ is a poset category and $p\in P$, then $P/p\cong\downarrow (p)$ the slice category $P/p$ is just the "principal ideal" $\downarrow(p)$ of elements $q\in P$ with $q\leq p$.

I have a few questions:

  1. What is "principal ideal" and why is it in quotes? Is that not an offical name?
  2. Why is "it" isomorphic to a slice category $P/p$? I can't find any inverses here...
  3. What do elements $q\in P$ have to do with anything if it's $\downarrow (p)$ and not $\downarrow(q)$?

(Probably question #3 is really silly, considering I don't understand 1 and 2.)

Any help appreciated, thank you.

P.S.: My math level: newbie

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1 Answer 1

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Possibly just put in quotes because it has not been defined previously. The definition is right there in the text.

An ideal in a poset is a set $I\subseteq P$ where if $y\in I$ and $x\leq y$ then $x\in I$. The definition of the principle ideal $\downarrow(p)$ is right there in the text. It is just the ideal $I=\{x:x\leq p\}$. $q$ is used above instead of $x$, and it is just used to define which elements of $p$ are in $\downarrow(p)$.

Now, the category corresponding to a poset $P$ has as objects the elements $x\in P$ and where $hom(x,y)$ has a single element if $x\leq y$ and is empty otherwise.

An ideal, $I$, is a subset of $P$, so it is also a poset, so it has a corresponding category.

If $P$ is a poset and $p\in P$ then the category $P/p$ is just the set of arrows $q\to p$, which is equivalent to the set of $q\leq p$, which then is equivalent to the category corresponding to $\downarrow(p)$.

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Thanks. Sorry, not sure what the difference between ideal and principal ideal is: it seems to me like they are both sets of objects less than a given object from some set. –  drozzy Oct 26 '12 at 18:07
    
In some cases, all ideals are principal, but in some posets, there are ideals that are not equal to $\downarrow(p)$. –  Thomas Andrews Oct 26 '12 at 18:11
    
For example, if $P$ is the real numbers under the usual ordering, then there are the closed intervals $(-\infty,p]$ which correspond to $\downarrow(p)$ and there are the open intervals $(-\infty,p)$ which are ideals but are not principal. –  Thomas Andrews Oct 26 '12 at 18:13
    
Also, in general, all of $P$ is an ideal, but, unless $P$ has a unique supremum, $P$ is not equal to $\downarrow(p)$ for any $p\in P$ –  Thomas Andrews Oct 26 '12 at 18:15
    
Also, you don't really need to know what an ideal is to understand the above argument, you just apply the definition given for a "principal ideal" given above. –  Thomas Andrews Oct 26 '12 at 18:19

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