What is a lift?

Question

What exactly is a lift? I wanted to prove that for appropriately chosen topological groups $G$ we can show that the completion of $\widehat{G}$ is isomorphic to the inverse limit $$\lim_{\longleftarrow} G/G_n$$

I wasn't sure how to construct the map so I asked this question to which I got an answer but I don't understand it. Searching for "lift" or "inverse limit lift" has not helped and I was quite sure that if $[x] \in \widehat{G}$ I could just "project" $x_n$ to $$\lim_{\longleftarrow} G/G_n$$ using the map $\prod_{n=1}^\infty \pi_n (x) = \prod_{n=1}^\infty x_n$ in $G/G_n$. Would someone help me finish this construction and explain to me what a lift is? Thank you.

Edit

I don't know any category theory. An explanation understandable by someone not knowing any category theory would be very kind.

In my answer I addressed the question 'what is a lift' in general. I think questions about this particular case (completion of a topological group) should be asked in the thread you mention. — wildildildlife, Jul 29 '12 at 12:21
@wildildildlife No, the person who answered is on leave so no one is going to look there and see my comments. — Rudy the Reindeer, Jul 29 '12 at 16:20
Given ƒ: X→Z and g₂: Y→Z, we ask for g₁: X→Z that factors through g₂. — isomorphismes, Sep 13 '15 at 18:59

score 8 · Answer 1 · answered Jul 29 '12 at 17:03

The term lift is merely meant to mean the following. Given a surjective map $G\to G'$ a lift of an element $x\in G'$ is a choice of $y\in G$ such that $y\mapsto x$ under this map.

In the linked answer you have $(x_n)\in \lim G/G_n$. Thus the notation has $x_n\in G/G_n$. The elements $y_n$ are just chosen preimages of $x_n$ under the natural surjection $G\to G/G_n$.

wildildildlife · Accepted Answer · 2012-07-29T18:58:37.967

4

See the Wikipedia article on lift:

In a category $C$, let be given two arrows $f,g$ with codomain $Z$. A lift (of $f$ with respect to $g$) is an arrow $F$ such that $g\circ F =f$. So it is a morphism from $f$ to $g$ in the over category $C/Z$.

Motivation: the idea is that $g:X\to Z$ indicates the vertical direction pointing down: $X$ lies above $Z$ via $g$. Think of $g$ as a fibre bundle, or covering map, or vertical projction. So a given $f:Y\to Z$ takes values in the lower space; and to lift $f$ (via $g$) means to 'let it take values in the upper space': find a 'modified' $F:Y\to X$ such that, after vertically projecting its values, we get $f$ back; in other words $p\circ F=f$. If we take $Y=[0,1]$ this is path-lifting.

In particular:

a lift of the identity $id:Z\to Z$ (via $g:X\to Z$) is an arrow $s:Z\to X$ such that $g\circ s=id$, i.e. a section of $g$.

(I believe in your case this is meant: a section of a quotient map $q:X\to X/\sim$ is a map $s:X/\sim\to X$ such that $s[x]\sim x$. One could say that $s[x]\in X$ is a lift of $[x]$.)

DUALLY: let be given two arrows $f,g$ with domain $Z$. A lift of $f$ (via $g$) is an arrow $F$ such that $F\circ f=g$. So it is a morphism from $f$ to $g$ in the under category $Z/C$.

Motivation: again the idea is that $g:Z\to X$ indicates the vertical direction, but now pointing up: $X$ lies above $Z$ via $g$. Think of $g$ as an inclusion; then:

a lift of $f:Y\to X$ (via an inclusion $g:Z\to X$) is an arrow $F:Y\to X$ such that $F\circ i=g$, i.e. an extension of $f$.

Related concepts are injective and projective objects. Moreover note that lifts need not exist in general (see also Zhen Lin's comment).

edited Jul 29 '12 at 18:58

answered Jul 29 '12 at 12:17

wildildildlife

5,550

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You should probably mention that, in most cases, the lift lives in the category of sets. For example, the identity map $\mathbb{Z} / p \mathbb{Z} \to \mathbb{Z} / p \mathbb{Z}$ has no lift with respect to the quotient map $\mathbb{Z} \to \mathbb{Z} / p \mathbb{Z}$ in the category of groups. – Zhen Lin Jul 29 '12 at 12:40
What is $C/Z$ and what is $Z/C$? – Rudy the Reindeer Jul 29 '12 at 16:28
Also, if $f: A \to B$ and $g : C \to B$, shouldn't a lift of $f$ be a map that lifts $f$ to $C$, i.e. a map $h: C \to A$ such that $fh = g$? – Rudy the Reindeer Jul 29 '12 at 16:30
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@ClarkKent: I have added some details, which I hope answer your questions. – wildildildlife Jul 29 '12 at 18:59
Nice, thank you very much! – Rudy the Reindeer Jul 29 '12 at 19:37
@wildildildlife: The dual of a lift is usually called an extension, whether or not $g$ is an inclusion. – Zhen Lin Jul 30 '12 at 04:55
So a lift does not have to be continuous? Most definitions on the internet speak of lifts of continuous maps. – Rudy the Reindeer Aug 22 '12 at 15:01
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Perhaps if the category is $\mathbf{Top}$, the arrows are continuous maps? – Rudy the Reindeer Aug 22 '12 at 15:04

What is a lift?

2 Answers2

Linked