What does it mean if a map corresponds to a path?

Question

I am solving an exercise and they wrote the following:

Let $f,g:S^1 \rightarrow U\subset \mathbb{R}^2$ be continous maps corresponding to paths $\gamma_f,\gamma_g:[0,1]\rightarrow U$

I somehow don't understand what this means. So we know that $$f,g:\{(x,y): ||(x,y)||=1\} \rightarrow U,\,\,\,\,\,\,\, (x,y)\mapsto f,g(x,y)$$ but how is $\gamma_f,\gamma_g$ defined?

Could someone explain this to me?

Thank you a lot.

so it's the introduction to an exercise we got from our prof. So afterwards we need to show something about homotopies, but I can't start with the proof since I don't understand $\gamma_{f,g}$ — user1294729, Nov 23 '21 at 09:26
I answered your question, but I cannot imagine that this point was not explained in a lecture. — Paul Frost, Nov 23 '21 at 09:31

Paul Frost · Accepted Answer · 2021-11-23T09:59:04.213

1

The circle $S^1$ is obtained as a quotient from $I = [0,1]$ by identifying the points $0$ and $1$. You can take $p : I \to S^1, p(t) = (\cos(2\pi t), \sin(2\pi t))$ as a quotient map. Then $\gamma_f = f \circ p, \gamma_g = g \circ p$.

To obtain an assignment $f \mapsto \gamma_f$ it would be sufficient to take any map $p : I \to S^1$. But to obtain a closed path $\gamma_f$ for each $f$, we need the minimal requirement that $p$ is a closed path in $S^1$, i.e. $p(0) = p(1)$. This is the case for our above $p$. But what is the relevance of our $p$ being a quotient map?

Given a closed path $\delta : I \to X$, i.e. with $\delta(0) = \delta(1)$, we see that there exists a unique function $\bar \delta : S^1 \to X$ such that $\bar \delta \circ p = \delta$. Since $p$ is a quotient map, this map is continuous (universal property of quotient maps). Thus we get an assigment $\delta \mapsto \bar \delta $. This is easily seen to be inverse to $f \mapsto \gamma_f$ .

edited Nov 23 '21 at 09:59

answered Nov 23 '21 at 09:29

Paul Frost

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sorry why do we need the quotient, so wouldn't p be enough to have, i.e. I don't see exaclty where the quotient comes in – user1294729 Nov 23 '21 at 09:32
Could you maybe explain this point, also which equivalence relation we need? – user1294729 Nov 23 '21 at 09:49
The equivalence relation on $I$ is given by identifying the points $0$ and $1$. Thus explicitly $s \sim t$ if $s = t$ or ${s,t} = {0,1}$. – Paul Frost Nov 23 '21 at 09:54
ah okey thank you! – user1294729 Nov 23 '21 at 09:56
There is minor point: You have to know that – Paul Frost Nov 23 '21 at 09:59
There is a minor point: You have to know that 1. all $p (t) = (\cos(2\pi t), \sin(2\pi t)) \in S^1$ (this is obvious), 2. $p(0) = p(1)$ (follows from the periodicity of $\cos, \sin$), 3. $p(I) = S^1$ and $p(s) \ne p(t)$ unless ${s,t} = {0,1}$. – Paul Frost Nov 23 '21 at 10:04
Sorry but isn't $p(0)=p(1)$ only because in $I$ $0\sim 1$ – user1294729 Nov 23 '21 at 10:07
i.e. they are in the same equivalence class. and since $p:[0,1] /_\sim \rightarrow S^1$ we have that $p(0)=p(1)$? – user1294729 Nov 23 '21 at 10:13

What does it mean if a map corresponds to a path?

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