- #1
trixitium
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I'm reading the first chapter of Topology by Munkres. There we can see:
"if [itex] f [/itex] is bijective, there exists a function from B to A called the inverse of [itex] f [/itex].
(...)
As another situation where care is needed, we note that it is not in general true that
[itex] f^{-1}(f(A_0) = A_0 [/itex] and [itex] f(f^{-1}(B_0)) = B_0 [/itex]. The relevant rules, which we leave you to check, are the following: If [itex] f: A \rightarrow B [/itex] and [itex] A_0 \subset A [/itex] and [itex] B_0 \subset B [/itex], then
[itex] A_0 \subset f^{-1}(f(A_0)) [/itex] and [itex] f(f^{-1}(B_0) \subset B_0 [/itex]
The first inclusion is equality if [itex] f [/itex] is injective and the second inclusion is equality if [itex] f [/itex] is surjective."
Are there any sense in talking about inverse considering that [itex] f [/itex] is not injective or surjective?
"if [itex] f [/itex] is bijective, there exists a function from B to A called the inverse of [itex] f [/itex].
(...)
As another situation where care is needed, we note that it is not in general true that
[itex] f^{-1}(f(A_0) = A_0 [/itex] and [itex] f(f^{-1}(B_0)) = B_0 [/itex]. The relevant rules, which we leave you to check, are the following: If [itex] f: A \rightarrow B [/itex] and [itex] A_0 \subset A [/itex] and [itex] B_0 \subset B [/itex], then
[itex] A_0 \subset f^{-1}(f(A_0)) [/itex] and [itex] f(f^{-1}(B_0) \subset B_0 [/itex]
The first inclusion is equality if [itex] f [/itex] is injective and the second inclusion is equality if [itex] f [/itex] is surjective."
Are there any sense in talking about inverse considering that [itex] f [/itex] is not injective or surjective?