Math Problem Statement

Let f: A → B, g: B → C, and h: C → D. Prove the following: i. If f and g are one-to-one, then g ∘ f is one-to-one. ii. If f and g are onto, then g ∘ f is onto. iii. (h ∘ g) ∘ f = h ∘ (g ∘ f).

Solution

Let's solve the problem step by step. We are given three functions f:ABf: A \to B, g:BCg: B \to C, and h:CDh: C \to D. We need to prove three statements based on the composition of these functions.

1. Prove: If ff and gg are one-to-one, then gfg \circ f is one-to-one.

Proof:

  • Let ff and gg be injective (one-to-one functions).
  • To prove that gfg \circ f is injective, assume g(f(x1))=g(f(x2))g(f(x_1)) = g(f(x_2)) for some x1,x2Ax_1, x_2 \in A.
  • Since gg is injective, we can conclude that f(x1)=f(x2)f(x_1) = f(x_2).
  • Since ff is injective, x1=x2x_1 = x_2.
  • Thus, gfg \circ f is injective (one-to-one).

2. Prove: If ff and gg are onto, then gfg \circ f is onto.

Proof:

  • Let ff and gg be surjective (onto functions).
  • To prove that gfg \circ f is surjective, take any cCc \in C. Since gg is surjective, there exists some bBb \in B such that g(b)=cg(b) = c.
  • Since ff is surjective, there exists some aAa \in A such that f(a)=bf(a) = b.
  • Therefore, g(f(a))=g(b)=cg(f(a)) = g(b) = c, which shows that for every cCc \in C, there exists an aAa \in A such that g(f(a))=cg(f(a)) = c.
  • Thus, gfg \circ f is surjective (onto).

3. Prove: (hg)f=h(gf)(h \circ g) \circ f = h \circ (g \circ f).

Proof:

  • We need to show that (hg)f=h(gf)(h \circ g) \circ f = h \circ (g \circ f).
  • By the definition of composition of functions, we have:
    • (hg)f(x)=h(g(f(x)))(h \circ g) \circ f (x) = h(g(f(x))),
    • h(gf)(x)=h(g(f(x)))h \circ (g \circ f)(x) = h(g(f(x))).
  • Since both sides simplify to h(g(f(x)))h(g(f(x))), the equality holds.
  • Therefore, (hg)f=h(gf)(h \circ g) \circ f = h \circ (g \circ f).

These are the required proofs for the given statements.


Do you want more details on any of the steps or have any questions?

Here are some related questions for further practice:

  1. Can you prove that the composition of two surjective functions is surjective?
  2. How would you define the inverse of a composition of two injective functions?
  3. What is the general property of function composition with respect to associativity?
  4. If gg is not surjective but ff is, is gfg \circ f surjective?
  5. Can a function be injective but not surjective? Provide an example.

Tip: Always pay attention to how function properties (injectivity and surjectivity) transfer through composition, as these are fundamental in proving larger properties of composed functions.

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Math Problem Analysis

Mathematical Concepts

Function Composition
Injectivity (One-to-One)
Surjectivity (Onto)
Associativity of Functions

Formulas

(g ∘ f)(x) = g(f(x))
(h ∘ g) ∘ f = h ∘ (g ∘ f)

Theorems

Theorem: Composition of injective functions is injective
Theorem: Composition of surjective functions is surjective
Associativity of function composition

Suitable Grade Level

Grades 11-12 or University-level