Math Problem Statement

Let f : A −→ B be a function. Show that it is not always the case that

f (f −1(Y )) = Y for all Y ⊆ B, by giving an explicit counterexample, but that the

equation is always true if f is surjective.

Solution

To show that f(f1(Y))=Yf(f^{-1}(Y)) = Y does not always hold for all subsets YBY \subseteq B, we can present a counterexample. Afterward, we'll demonstrate that this equation holds if ff is surjective.

Step 1: Counterexample

Let:

  • A={1,2}A = \{1, 2\}
  • B={a,b,c}B = \{a, b, c\}
  • Define a function f:ABf: A \to B by: f(1)=aandf(2)=bf(1) = a \quad \text{and} \quad f(2) = b

Here, ff is not surjective because there is an element cBc \in B that is not in the range of ff.

Now, take the subset Y={c}BY = \{c\} \subseteq B.

  1. Compute f1(Y)f^{-1}(Y): Since f(1)=af(1) = a and f(2)=bf(2) = b, neither of these values is cc. Therefore, no element in AA maps to cc, and thus: f1(Y)=f1({c})=f^{-1}(Y) = f^{-1}(\{c\}) = \emptyset

  2. Compute f(f1(Y))f(f^{-1}(Y)): Since f1(Y)=f^{-1}(Y) = \emptyset, applying ff to \emptyset yields: f(f1(Y))=f()=f(f^{-1}(Y)) = f(\emptyset) = \emptyset

  3. Compare f(f1(Y))f(f^{-1}(Y)) and YY: Here, f(f1(Y))={c}=Yf(f^{-1}(Y)) = \emptyset \neq \{c\} = Y.

Thus, we have a case where f(f1(Y))Yf(f^{-1}(Y)) \neq Y, which confirms that the equation does not always hold for arbitrary functions.

Step 2: Proof for Surjective Functions

Suppose f:ABf: A \to B is surjective, meaning that every element of BB has a preimage in AA. We want to show that for any subset YBY \subseteq B, f(f1(Y))=Yf(f^{-1}(Y)) = Y.

  1. Show f(f1(Y))Yf(f^{-1}(Y)) \subseteq Y:

    • For any yf(f1(Y))y \in f(f^{-1}(Y)), there exists some xf1(Y)x \in f^{-1}(Y) such that f(x)=yf(x) = y.
    • Since xf1(Y)x \in f^{-1}(Y), we have f(x)Yf(x) \in Y.
    • Therefore, yYy \in Y, which proves f(f1(Y))Yf(f^{-1}(Y)) \subseteq Y.
  2. Show Yf(f1(Y))Y \subseteq f(f^{-1}(Y)):

    • Since ff is surjective, for each yYBy \in Y \subseteq B, there exists an xAx \in A such that f(x)=yf(x) = y.
    • By the definition of f1(Y)f^{-1}(Y), this xx is in f1(Y)f^{-1}(Y).
    • Thus, y=f(x)f(f1(Y))y = f(x) \in f(f^{-1}(Y)).
    • This shows Yf(f1(Y))Y \subseteq f(f^{-1}(Y)).

Since both inclusions hold, we conclude f(f1(Y))=Yf(f^{-1}(Y)) = Y for all YBY \subseteq B when ff is surjective.

Summary

  • Counterexample: When ff is not surjective, we found f(f1(Y))Yf(f^{-1}(Y)) \neq Y.
  • Proof for Surjectivity: If ff is surjective, f(f1(Y))=Yf(f^{-1}(Y)) = Y for any subset YBY \subseteq B.

Would you like further details or have any questions?


Related Questions

  1. How does the injectivity of ff affect the equation f(f1(Y))=Yf(f^{-1}(Y)) = Y?
  2. Can f(f1(Y))=Yf(f^{-1}(Y)) = Y still hold if ff is injective but not surjective?
  3. What are the implications of f1(f(X))=Xf^{-1}(f(X)) = X for subsets XAX \subseteq A when ff is injective?
  4. How would the equation change if ff were bijective?
  5. How can the preimage f1(Y)f^{-1}(Y) be used in probability theory or set theory?

Tip

When working with preimages and images in functions, carefully consider whether the function is surjective, as this often influences whether properties like f(f1(Y))=Yf(f^{-1}(Y)) = Y hold.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Functions
Surjectivity
Preimages and Images

Formulas

f(f^{-1}(Y))

Theorems

Properties of Surjective Functions

Suitable Grade Level

Undergraduate Mathematics