   Chapter 1.5, Problem 10E

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# Let f and g be mappings from Α to Α . Prove that if f ∘ g is invertible, then f is onto and g is one-to-one.

To determine

To prove: fg is invertible, then f is onto and g is one-to-one.

Explanation

Given information:

f and g are mappings from A to A.

Formula used:

1) Definition: Let f:AB. Then f is called onto or subjective, if and only if B=f(A).

2) Definition: Let f:AB. Then f is called one-to-one or injective, if and only if different elements of A always have different images under f.

3) Definition: Let g:AB and f:BC. The composite mapping fg is the mapping from A to C defined by (fg)(x)=f(g(x)) for all xA.

Proof:

Let f and g be mappings from A to A and the mapping fg:AA is invertible.

To show that f is onto and g is one-to-one.

Since fg is invertible, there exists a mapping h:AA such that

(fg)h=IA and h(fg)=IA where IA:AA is an identity mapping.

First prove that f is onto.

Let aA; then ((fg)h)(a)=IA(a)

But IA(a)=a

Therefore, ((f

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