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 April 13th, 2014, 05:34 PM #1 Newbie   Joined: Apr 2014 From: Colombia Posts: 1 Thanks: 0 a simple question on functions Hey everyone, I'm in my first semester of undergraduate mathematics and have a simple question on functions: In the book we are using for the course on Introductory Maths there is a Theorem that says: If f is a function such that f: A --> B, then f is one to one if and only if f is invertible. To me the theorem makes perfect sense but then there is a statement : "Note that if f is invertible, then (f^-1)^-1 = f, and so f^-1 is invertible and also one to one by the previous theorem". Why we can conclude that f^-1 is invertible and that its inverse equals f only with this information? In my opinion a Lema is needed What do you guys think? TY
 April 14th, 2014, 02:16 AM #2 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra I think the statement is okay. Just have to be careful: while $f$ is a function from $A$ to $B$, $f^{-1}$ (provided $f$ is injective) is a function from $f(A)$ (the range of $f$) to $A$. This is because unless $f$ is surjective there will be some elements in $B$ not in the range of $f$ and hence not in the domain of $f^{-1}$.
April 14th, 2014, 03:16 PM   #3
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Quote:
 Originally Posted by DanielR If f is a function such that f: A --> B, then f is one to one if and only if f is invertible. To me the theorem makes perfect sense but then there is a statement : "Note that if f is invertible, then (f^-1)^-1 = f, and so f^-1 is invertible and also one to one by the previous theorem".
There's some ambiguity here. Two ambiguities in fact.

* A one-to-one function is a function that hits every target at most once. That's also known as an injection. The function f: ℕ -> ℕ given by f(x) = 2x is an injection, because every element of the range gets hit at most once. But some targets don't get hit at all. For example the number 5 does not get hit (nor does any other odd number). In this case, f is not invertible.

A bijection, also known as a one-to-one correspondence, is an injection that's also a surjection, or "onto" function. Everything in the range gets hit exactly once. A bijection is always an invertible function. And that's what I think you mean. The ambiguity is between one-to-one function and one-to-one correspondence. That's why the words injection and bijection are preferable. There's no ambiguity in those.

* The other ambiguity is with the inverse of a function. As Olinguito points out, an injection does has an inverse, but it's a different kind of inverse. This kind of inverse maps subsets of the range to subsets of the domain. So the function f(x) = 2x on the naturals does not have a functional inverse; but it does have a subset-type inverse. For example the inverse image of the odd numbers is the empty set; and the inverse image of the even numbers is all of ℕ.

Unfortunately the word "inverse" is used with both meanings, and this invariably causes confusion for students.

So: A bijection f:A -> B always has a functional inverse g:B->A.

Every function f:A->B always has a subset inverse g:P(B) -> P(A) where P(X) denotes the power set of X, which is the collection of all subsets of X. There's nothing special about injections or even bijections in this context. You can always take a subset of the range and ask what subset of the domain hits it.

Can you say which meaning of inverse you intended; and which use of "one-to-one" you intended? They both make a difference.

When you say "Note that if f is invertible, then (f^-1)^-1 = f, and so f^-1 is invertible and also one to one by the previous theorem" that indicates to me than you mean to say that a bijection always has a functional inverse. That's a true statement; but the phrase "one-to-one function" is generally taken to mean an injection, not necessarily a bijection, which makes your statement false.

Takeaway: Use the terminology injection/bijection/surjection; and work through the two meanings of inverse.

Last edited by Maschke; April 14th, 2014 at 04:10 PM.

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