how do you tell if a specific function has an inverse or not?

how do you tell if a specific function has an inverse or not?
Er, by showing it's a bijection. But one needs to know what the function IS.
Look, this comes up from timetotime here, and (I believe) confusion is caused by some horrid, but conventional, notation. So let's step through it.
First, without any rigour, define an injection by: for any there is at most one such that .
Now, with the same lack of rigour define a surjection as: for any there is at least one such that .
Roll these together, and you have the bijection: for any there is at least one and at most one such that .
Now. The "horrid" notation I referred to above arises as follows: no function, on sets or otherwise, is allowed to have multiple images (these being the result of applying our function to its domain, here ). But, by our surjection above, the preimage MUST be a nonempty set, say .
The preimage of our injection is a set also. So, by my above, for the injection , then either or . This called a "singleton set"
For the bijection we will therefore have that .
In this particular circumstance is customary to ever so slightly abuse the notation, and to equate and by a further slight abuse to call the inverse of
What Guitarist said is correct. But perhaps a bit difficult to follow if you have not seen this before.Originally Posted by Heinsbergrelatz
In simpler language, f is invertible if given f(x) you can figure out what x is.
Now, strictly speaking this only really means that f is injective or onetoone, but any injective function is invertible if you restrict the range to the image thereby making the function surjective by fiat.
so basically e.g should equal
Well, you need to specify your function. So, for example, if , then yes, this true.
But if, as is equally possible without this specification, that then it is false, since and also then the preimage .
This last is a SET, as I tried to explain. DrRocket suggested you may not understand my post; I found it hard to believe at the time, but now I see he must be right.
the example i gave was where the function has an inverse. then more generally speaking;
a function defined should equal to in order for it to have an inverse?
i didnt really read your post, (no offense) because such terms mentioned by your post, i didnt quite yet encounter. therefore, instead i read Dr. Rockets explanation which was alittle more concise and easier to understand. but now you said that "quoted by me", i do understand what you are trying to explain.This last is a SET, as I tried to explain. DrRocket suggested you may not understand my post; I found it hard to believe at the time, but now I see he must be right.
Yes, GREAT offense, as it seems I wasted my precious timeOriginally Posted by HeinsbergrelatzThen you only had to ask, and we could get a discussion going which may been of interest to ALL our readers.because such terms mentioned by your post, i didnt quite yet encounter.
sorry for making you waste your time, but i did read it eventually, though not all terms i could understand. But usually i seldom read posts that i hardly understand, or not yet learned. just a habit i guess, but since it offended you , my apologies.Yes, GREAT offense, as it seems I wasted my precious time
yes, that would have been much better. too badThen you only had to ask, and we could get a discussion going which may been of interest to ALL our readers.
If it passes the horizontal line test, than there exists an inverse.Originally Posted by Heinsbergrelatz
That works if the function is a realvalued function of a real variable.Originally Posted by Ellatha
I read this, so you certainly didn't waste your time. Would you mind going through it with a little more rigor? The bijection, injection, surjection, and all the like? Notation and all if you could, I'm not well versed in the notation, but I can follow the explanations so far easily.Originally Posted by Guitarist
injection  aka oneoneOriginally Posted by Arcane_Mathematician
surjection  aka onto
bijection  onetoone and onto
And with two sets, say X and Y, a function will be an injection so long as there is, for every at most on such that . does that mean that there can be no x, such that ? is there a requirement that at least one exists such that for one ? Does that make any sense?
You have the definition of an injection (onetoone function) correct.Originally Posted by Arcane_Mathematician
An injection need not be onto, so there may be a y that is not f(x) for any x.
If every y is f(x) for some x then f is "onto" or "surjective" (same thing).
If f is both an injection and a surjection then it is a bijection. A bijection and an invertible function are the same thing.
Well I am not sure I can add too much to my previous, except perhaps this.Originally Posted by Arcane_Mathematician
It essential to specify what sort of objects our function takes as input and what sort of objects it returns as output; these are known respectively as the domain and the codomain of our function.
Take the simple case that , the integers, where, say, . This is obviously an injection, since 3 is obviously integer, yet there is no such that . One says that 3 is not in the range of our function.
But if I choose another codomain, say the even numbers (those integers exactly divisible by 2), I may have (this is standard notation for the even integers) is a bijection; one half of every even integer is an integer, even or odd. But notice that , so we have that an injection from a set to a set can be a bijection from a set to a subset.
Or consider another function with the same domain and codomain, say , This is clearly a surjection, as for any integer. But again, by changing the domain to the nonnegative integers, call them we may recover a bijection.
Again, a surjection from a set set to a set may be a bijection from a subset to a set. I hope you can see the symmetry here.
In fact, if we have some reason (we often do) to retain our original domain, and by noting that we may have what is called the restriction of the function to this subset of the domain, which is usually written , which in the case in point is our bijection
As far as I am aware this is not used all that often, but there are cases from trigonometry where it is very widely used. Can you think what they are?
the functions arcsine and arccosine pop into mind immediately, where the domain is restricted only to one half of a cycle for each of the respective functions. I imagine it would also be used for the inverse of tangent, though less obviously.
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