# Uni homework - bijection

Tjaz Gantar
Homework Statement:
Find a bijection between A and B.
Relevant Equations:
A = (−∞, 0) ∪ (0, ∞) ⊂ R,
B = {(x, y) ∈ R2| x2 − y2 = 1}
So I don't really have any goods ideas on how to try and solve this. I only know that using tangent function is supposedly a good idea.

archaic
I think that if you isolate ##y##, you can make a piecewise injective function of ##x## that reaches all values of ##A##.

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Homework Statement: Find a bijection between A and B.
Homework Equations: A = (−∞, 0) ∪ (0, ∞) ⊂ R,
B = {(x, y) ∈ R2| x2 − y2 = 1}

So I don't really have any goods ideas on how to try and solve this. I only know that using tangent function is supposedly a good idea.

You have ##\sec^2 t - \tan^2 t = 1## and ##\cosh^2 t - \sinh^2 t = 1##.

Either of these might be a good starting point.

Have you drawn a graph of B?

archaic
You have ##\sec^2 t - \tan^2 t = 1## and ##\cosh^2 t - \sinh^2 t = 1##.

Either of these might be a good starting point.

Have you drawn a graph of B?
Can't we consider ##f(x) := \sqrt{x^2-1}## for ##x>1## and ##-\sqrt{x^2-1}## for ##x<-1##?

Delta2
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Can't we consider ##f(x) := \sqrt{x^2-1}## for ##x>1## and ##-\sqrt{x^2-1}## for ##x<-1##?

Okay, where are you mapping the following 4 points in B?

##x = 2, y = \sqrt{3}; \ x = 2, y = -\sqrt{3}; \ x = -2, y = \sqrt{3}; \ x = -2, y = -\sqrt{3}##

archaic
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Doesn't a bijection mean that the function is both injective and surjective?

Yes. I note that this is not your homework!

B is not the graph of a function.

SammyS and archaic
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Homework Statement: Find a bijection between A and B.
Homework Equations: A = (−∞, 0) ∪ (0, ∞) ⊂ R,
B = {(x, y) ∈ R2| x2 − y2 = 1}

So I don't really have any goods ideas on how to try and solve this. I only know that using tangent function is supposedly a good idea.
Hello @Tjaz Gantar .

I see that although you have been a member for more than half a year, this is your first post here at PF.

As @PeroK suggests, draw a graph of B.

By the way; do you mean to say x and y are squared in the definition of set B as in the following?
B = {(x, y) ∈ ℝ2| x2 − y2 = 1}​
.

PeroK
archaic
Okay, where are you mapping the following 4 points in B?

##x = 2, y = \sqrt{3}; \ x = 2, y = -\sqrt{3}; \ x = -2, y = \sqrt{3}; \ x = -2, y = -\sqrt{3}##
Hey, I hope I'm not intruding on the HW further.
Isn't it irrelevant whether we map every element of the starting set if every element of the final has an antecedent? The function I gave takes all ##x##s in ##B## (##x\in\mathbb{R}-[-1,1]##) and gives you ##A##.

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Hey, I hope I'm not intruding on the HW further.
Isn't it irrelevant whether we map every element of the starting set if every element of the final has an antecedent? The function I gave takes all ##x##s in ##B## (##x\in\mathbb{R}-[-1,1]##) and gives you ##A##.

I don't know what that means.

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Hey, I hope I'm not intruding on the HW further.
Isn't it irrelevant whether we map every element of the starting set if every element of the final has an antecedent? The function I gave takes all ##x##s in ##B## (##x\in\mathbb{R}-[-1,1]##) and gives you ##A##.
Yes, it's relevant. You need the map to be both one to one and onto. a bijection.

archaic
archaic
I don't know what that means.
Yes, it's relevant. You need the map to be both one to one and onto. a bijection.
I thought I remembered the definition correctly, my bad!

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I thought I remembered the definition correctly, my bad!

Take an example where ##B## is defined by ##x^2 + y^2 = 1##. Then we can define an injection from ##[0, 2\pi)## to ##B## using ##x = \cos t, \ y = \sin t##.

If ##B## were one half of a hyperbola, then the problem would be equally simple. But, ##B## is the two disjoint halves of the hyperbola. That makes things a little trickier.

But note that ##A## also is disjoint.

Another idea that may be useful is that the composition of two injections is a injection.

archaic
archaic
Take an example where ##B## is defined by ##x^2 + y^2 = 1##. Then we can define an injection from ##[0, 2\pi)## to ##B## using ##x = \cos t, \ y = \sin t##.

If ##B## were one half of a hyperbola, then the problem would be equally simple. But, ##B## is the two disjoint halves of the hyperbola. That makes things a little trickier.

But note that ##A## also is disjoint.

Another idea that may be useful is that the composition of two injections is a injection.
Are we trying to go from ##A## to ##B##?

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Are we trying to go from ##A## to ##B##?

A bijection, by definition, goes both ways.

Last edited:
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Are we trying to go from ##A## to ##B##?
Actually, there is an injection from ##B## to ##A## which is quite remarkable. I'm not sure what the simplest way to find it is. It's the inverse of the one I found going from ##A## to ##B## starting with the hyperbolic trig parameterisation of ##B##.

It's definition is absurdly simple and looks unbelievable.

sysprog
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An injection, by definition, goes both ways.
Presumably you meant to say: a bijection, by definition, goes both ways.

PeroK
Antarres
As @PeroK pointed out earlier, if we look at the graph of the second set, we find that it is a hyperbola. So, we see hyperbola has two branches, and our set A, also has two 'branches', that is, two disjunct intervals. Also, both of these are in a sense symmetrical. Hyperbola is symmetrical with respect to ##y##-axis, while set A is symmetrical with respect to 0. So the idea could be to create parametrization of each branch of the hyperbola onto one of those intervals. By symmetry, the other branch could be parametrised trivially after this has been done. So you're searching for a function that maps positive real numbers into the positive(##x>0##) branch of the hyperbola, for example.

Tjaz Gantar
Hello @Tjaz Gantar .

I see that although you have been a member for more than half a year, this is your first post here at PF.

As @PeroK suggests, draw a graph of B.

By the way; do you mean to say x and y are squared in the definition of set B as in the following?
B = {(x, y) ∈ ℝ2| x2 − y2 = 1}​
.
Well I understand bijection and know what i need to do, I just had no idea how to make it so that it really is a bijection if you understand what i mean by that:D
And yes, its ment to be squared, didn't notice that it was written like that

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Well I understand bijection and know what i need to do, I just had no idea how to make it so that it really is a bijection if you understand what i mean by that:D
And yes, its ment to be squared, didn't notice that it was written like that

I don't know if there is an easier way to get a solution, but what I did was:

Start with the parameterisation of ##B## using the hyperbolic trig functions. Then, remember that the composition of two bijections is a bijection.

Tjaz Gantar
As @PeroK pointed out earlier, if we look at the graph of the second set, we find that it is a hyperbola. So, we see hyperbola has two branches, and our set A, also has two 'branches', that is, two disjunct intervals. Also, both of these are in a sense symmetrical. Hyperbola is symmetrical with respect to ##y##-axis, while set A is symmetrical with respect to 0. So the idea could be to create parametrization of each branch of the hyperbola onto one of those intervals. By symmetry, the other branch could be parametrised trivially after this has been done. So you're searching for a function that maps positive real numbers into the positive(##x>0##) branch of the hyperbola, for example.
So if i make it
x=t (t being element of A), y=|√(x2 - 1)| for t≥1
and
y=tan(π(t-1)/2), x=|√(y2+1)| for 0<t<1
would that be okay from "the right part" of A to "the right part" of the hyperbola?

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So if i make it
x=t (t being element of A), y=|√(x2 - 1)| for t≥1
and
y=tan(π(t-1)/2), x=|√(y2+1)| for 0<t<1
would that be okay from "the right part" of A to "the right part" of the hyperbola?

Yes, that looks right.

Note that ##\sqrt{x}## is non-negative, by definition.

Tjaz Gantar
Tjaz Gantar
Yes, that looks right.
okay its nice to see progress in my head hah thanks

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In theory we can also use Schroeder Bernstein and produce injections from each set to the other. Not likely but it may have been used in the class without proof.

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