Proving the Composition of Two Isomorphisms is an Isomorphism

In the context of mathematical structures such as groups, rings, or vector spaces, proving that the composition of two isomorphisms is an isomorphism is a fundamental concept. This article delves into the detailed proof and key definitions necessary to understand and prove this statement. By the end, you will have a comprehensive understanding of why the composition of two isomorphisms is itself an isomorphism.

Key Definitions

Before diving into the proof, it's essential to understand the definitions of an isomorphism and function composition.

Isomorphism

A function f: A to B is an isomorphism if it is a bijection (one-to-one and onto) and f preserves the structure of the elements in A and B. The structure can vary depending on the context, such as addition, multiplication, or other algebraic operations.

Function Composition

The composition of two functions f and g, denoted as g 'circ f, is defined by g 'circ f(x) g(f(x)) for all x in A.

Proof

To prove that the composition of two isomorphisms is also an isomorphism, we will break down the proof into several parts:

Bijectiveness

Firstly, we need to show that g 'circ f is bijective (i.e., both a bijection and an isomorphism).

Given that f: A to B and g: B to C are both isomorphisms, they satisfy the conditions for bijections:

f is a bijection: There exists an inverse function f-1: B to A such that: f-1(f(x)) x for all x in A f(f-1(y)) y for all y in B g is a bijection: There exists an inverse function g-1: C to B such that: g-1(g(y)) y for all y in B g(g-1(z)) z for all z in C

Composition of Inverses

We will now show that the composition of isomorphisms g 'circ f is also a bijection by finding its inverse:

Define (g 'circ f)-1 as g-1 'circ f-1. Verify this by checking: For any z in C: [ (g 'circ f)-1 'circ (g 'circ f) (x) f-1 'circ g-1 'circ g 'circ f (x) f-1 'circ g-1 'circ g (f(x)) f-1 'circ g-1 (y) f-1 (x) ] Note: y f(x) [ x ] for all x in A. Similarly, for z in C: [ (g 'circ f) 'circ (g-1 'circ f-1) (z) g 'circ f 'circ g-1 'circ f-1 (z) g 'circ f 'circ g-1 (y) g 'circ x z ] for all z in C, where y g-1 (z).

Structure Preservation

To show that g 'circ f preserves the structure, consider that:

If f and g are structure-preserving, i.e., f preserves the structure in A and B, and g preserves the structure in B and C, then their composition g 'circ f will also preserve the structure. For any a, b in A, we have: [ g(f(ab)) g(f(a) g(f(b))) ] (since g preserves the structure in B and C). [ g(f(a)) g(f(b)) ] (since f preserves the structure in A and B). [ (g 'circ f)(ab) ]

Conclusion

Since g 'circ f is bijective (by construction) with an inverse, and it preserves the structure, we conclude that g 'circ f is an isomorphism.

Thus, the composition of two isomorphisms is indeed an isomorphism, which is a fundamental concept in algebra and various mathematical structures.