Vtyjkyuk

So today in my Algebraic Topology class we were trying to construct a homotopy and at one point we needed to basically construct a (continuous) bijection between some intervals $[a, b]$ and $[c, d]$ sending $a mapsto c$ and $b mapsto d$.

My professor said that a bijection $varphi : [a, b] to [c, d]$ would be a linear function of the form $varphi(x) = alpha x + beta$ and adding the contraints that $varphi(a) = c$ and $varphi(b) = d$ we get a linear system of equations $$alpha cdot a + beta = c$$ $$alpha cdot b + beta =d.$$

Solving this system we get $$alpha = fracc-da-b$$ and $$beta=fracad-bca-b$$

so that $varphi$ is given by $$varphi(x) = left(fracc-da-bright)x + fracad-bca-b$$

Now I was not aware that we could construct a (continuous!) bijection between connected closed intervals of $mathbbR$ so easily. My question is how exactly did my professor know that such a (continuous) bijection would be a linear function of the above form? I've never seen any theorem of the sort or something similar written in any textbook.

Here are two pieces of knowledge which, if you possess them and put them together, will tell you that a linear formula like this must exist.

1. Given two closed intervals $I,J subset mathbb R$, there is a similarity transformation of the real line $f : mathbb R to mathbb R$ such that $f(I)=J$.

Just to be clear about the definition, to say that $f$ is a similarity transformation means that it stretches distances by a uniform amount, i.e. there is a constant $a>0$ such that $|f(x)-f(y)| = a |x-y|$. If you have ever studied Euclidean geometry then you might be familiar with similarity transformations, this is just the 1-dimensional version. Given $I$ and $J$ the constant $a$, of course, is just the length of $J$ divided by the length of $I$.

2. The similarity transformations of the real line are precisely the first degree polynomials, i.e. the transformations of the form $f(x)=alpha x + beta$ with coefficients $alpha in mathbb R - 0$ and $beta in mathbb R$.

This is not very hard to prove: if $f(x)=alpha x + beta$ prove that it is a similarity transformation with stretch constant $a = |alpha|$; and if $f(x)$ is a similarity transformation with stretch constant $a$ then set $beta = f(0)$, and prove that either $f(x)= a x + beta$ or $f(x) = -ax + beta$.

Think in stages:

i) $,f_1(x) = x-a$ takes $[a,b]$ to $[0,b-a].$

ii) $,f_2(x) = x/(b-a)$ takes $[0,b-a]$ to $[0,1].$

iii) $,f_3(x) = (d-c)x$ takes $[0,1]$ to $[0,d-c].$

iv) $,f_4(x) = x+c$ takes $[0,d-c]$ to $[c,d].$

The map $f_4circ f_3 circ f_2 circ f_1$ then does the job.

It should be intuitively clear you can "stretch" and "slide" any two interval to overlap.

So to "stretch" $[a,b]$ to be as long as $[c,d]$ you just have to multiply each of the $x in [a,b]$ by $frac textlength of [c,d]textlength of [a,b] = frac d-cb-a$ to get the point $x *frac d-cb-ain [a*frac d-cb-a, b*frac d-cb-a]$ which will have a length of $b*frac d-cb-a - a*frac d-cb-a= frac d-cb-a(b - a) = d-c$.

And to "slide" it, we need to slide the point $a*frac d-cb-a$ to $c$ so we need to add $c - a*frac d-cb-a$ so that the point $x *frac d-cb-a$ gets slid to $x *frac d-cb-a + (c - a*frac d-cb-a)=frac d-cb-ax + frac c(b-a) - a(d-c)b-a = frac c-db-ax + frac cb-adb-a$

So that's your bijection.

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Perhaps an even easier way to see it, is to take Randall's sugestion in the comments. On a $x,y$ cartesian plane look at the rectangular section of $[a,b]$ in the $x$ axis and $[c,d]$ in the $y$ access. Now just draw a line from one corner to the other. That line will be your bijection.

(I'll be honest. I never thought of this. It's.... trivially easy. But... I never came up with it on my own.)

(The slope of the line is $frac d-cb-a$ and it contains the point $(a,c)$ so the equation of the line is $y- c = frac d-cb-a(x-a)$ which becomes $y = f(x) = frac c-db-ax + frac cb-adb-a$)

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I suppose the way that is most into intuitive for me is:

A point $x in [a,b]$ is some proportion $t$ of the length of $[a,b]$ past $a$. That is $x = a + t*(b-a)$ for some real number $0 le t le 1$.

We want to transfer $x = a + t(b-a)$ to a point that is the same proportion of of the length of $[c,d]$ past $c$. That is we want to map $x = a+t(b-a) mapsto c + t(d-c)$.

A little bit of algebra and we get $t = frac x-ab-a$ and so we want to map $x mapsto c + frac x-ab-a(d-c) = frac d-cb-ax + (c - afrac d-cb-a)= frac c-db-ax + frac cb-adb-a$