Inverse of Function: Topology by Munkres Ch 1

[trixitium]

I'm reading the first chapter of Topology by Munkres. There we can see:

"if f is bijective, there exists a function from B to A called the inverse of f.

(...)

As another situation where care is needed, we note that it is not in general true that

f^{-1}(f(A_0) = A_0 and f(f^{-1}(B_0)) = B_0. The relevant rules, which we leave you to check, are the following: If f: A \rightarrow B and A_0 \subset A and B_0 \subset B, then

A_0 \subset f^{-1}(f(A_0)) and f(f^{-1}(B_0) \subset B_0

The first inclusion is equality if f is injective and the second inclusion is equality if f is surjective."

Are there any sense in talking about inverse considering that f is not injective or surjective?

 

[Latrace]

Ah, this is a common misunderstanding, due to a notation issue. If f : A \rightarrow B is any function and B_0 \subset B, then by f^{-1}(B_0) people always mean the set \{ x \in A | f(x) \in B_0 \}. Notice that this has meaning even if f is not a bijection. Likewise, if A_0 \subset A, we have f(A) = \{ f(x) | x \in A \}. I think that with these definitions you should be able to understand the conclusions made in your book.

 

[HallsofIvy]

Oh, dear, oh, dear! Just seeing this question makes me want to hide under the bed!

The very first time I had to give an explanation of a proof to a class in a topology class, it involved f^{-1}(A) for A a set and I did the whole thing assuming f was invertible!

If f is a function from set X to set Y, and A is a subset of X, B a subset of Y, then we define f(A) to be the set of all y in Y such that f(x)= y for some x in A and f^{-1}(B) to be the set of all x in X such that f(x) is in B.

IF f is "one to one and onto", that is, if f is invertibe, then we can show that f^{-1}(f(A))= A, but f does not have to be invertible, or even defined on set B for f^{-1}(B) to be defined.

For example, let f:R=>R be defined by f(x)= x² and let B= [-4, 4]. Then f^{-1}(B)= [-2, 2]. f(2)= f(-2)= 4 so both 2 and -2 are in f^{-1}(B) and for any x between -2 and 2, -4< 0< f(x)< 4, so x is also in f^{-1}(B). If x< -2 or x> 2, f(x)> 4 so not in [-4, 4].

Even f^{-1}([-4, -1]) is defined. Because there is NO x such that f(x)= x^2 is in [-4, -1] so f^{-1}([-4, -1]) is the empty set.