Let X be a non-empty set and f a mapping of X into itself. Show that f is one-to-one onto iff there exists a mapping g of X into itself such that fg = gf = iX. If there exists a mapping g with this property, then there is only one such mapping. Why?

Related BrainMass Solutions

• Let X and Y be non-empty sets and f a mapping of X into Y. Show that f is one-to-one iff there exists a mapping g of Y into X such that gf = iX.

  143556 Let X and Y be non-empty sets and f a mapping of X into Y. Show that f is one-to-one iff there exists a mapping g of Y into X such that gf = iX.

• Let X and Y be non-empty sets and f a mapping of X into Y. Show that f is onto iff there exists a mapping h of Y into X such that fh = iX.

  143557 Let X and Y be non-empty sets and f a mapping of X into Y. Show that f is onto iff there exists a mapping h of Y into X such that fh = iX.

• Topology and mapping functions

  X be a non-empty set.

• Discrete Mathematical Definitions

  Surjective (onto) A mapping/function f: X -> Y is surjective if for every value y in the set Y, there is at least one value x in X such that f(x) = y.

• Non-Empty Subsets of a Group

  This implies that x=e, and hence e∈H. It remains to show that together with any h the set H contains h^(-1). Since the above-defined mapping f is bijective and e∈H, as it was shown, there exists an element x in H such that hx=e.

• Aut(G), the set of automorphisms of G, is also a group.

  Herstein It is proven that if G is a group, then Aut(G), the set of automorphisms of G, is also a group. The solution is detailed and well presented.

• Automorphism of a group

  For every , there exists such that Therefore is one-one mapping of onto itself.

• finite Abelian group

  Then a mapping f: G -> H (read: a mapping f from G to H) is said to be a homomorphism iff: For all elements a, b in G, f(a.b) = f(a)*f(b) Properties: If e' is the identity of H, then f(e) = e' Let "^-1" denote the inverse of an element.

• Maximum Modulus Principle

  Theorem: Let f: D → D be a one-one analytic map of D onto itself and suppose f(z) = 0. Then there is a complex number c with |c| = 1 such that f = cϕa.