MAT102 Lecture 09B

• Images and Preimages
• Graphs in Math
• 3
  • 
  • It's injective only, as surjectivity would require that everything in the Codomain maps to something in the Domain.
  • Two elements from the domain mapping to one element in the codomain is a counterexample for injectivity.
  • Bijective is one-to-one
• 4
• 7
  • If we two outputs the same, inputs same
  • Injective
    • $f(a)=f(b)$
    • $(a^2,a^3)=(b^2,b^3)$
    • $a^2=b^2$
      • $\sqrt{a^2=b^2}$
      • $\pm a=\pm b$
      • Doesn't show us much
    • $a^3=b^3$
      • $\sqrt[3]{a^3=b^3}$
      • $a=b$
      • This doesn't have $\pm$
      • To prove this #tk Prove the cubic map is injective
      • $g:\mathbb{R}\to \mathbb{R}$ where $x\mapsto x^3$
        • $g(x)=g(y)$
        • WTS: $x=y$
        • $x^3=y^3$
        • $\sqrt[3]{x^3=y^3}$
        • $x=y$
      • $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
  • Surjective
    • Counterexample: Not surjective, as $(-1,x)$ is never reached. Since $k^2\ge 0$ always. Doesn't matter what $x$ is.
    • Other examples:
      • $(3,7)$
      • $(0,1)$
        • You can't have $k^2=0$ and $k^3=1$ with the same $k$.
    • See that $f(k)=(k^2,k^3)=(-1,x)$
    • Argue that $k^2\ne -1$
• 9
  • 
  • If one component function is injective, then the composition will be injective.
  • #tk generalize the proof for if $f$ is injective then $h$ is injective
  • Claim: If $h:A\to B\times B$ as $h(x)=(f(x),g(x))$ and either $f$ or $g$ is injective, then $h$ is injective
  • Proof:
    • Without loss of generality
      • $f:A\to B$ is injective
      • So $f(x)=f(y)⟹x=y$
      • Want to show that $h(x)=h(y)⟹x=y$
      • $h(x)=(f(x),g(x))$
      • $h(y)=(f(y),g(y))$
      • $(f(x),g(x))=(f(y),g(y))$
      • $f(x)=f(y)$
      • Since $f$ is injective, then $f(x)=f(y)⟹x=y$
      • Since we have one component of $h(x)=h(y)$ being injective, it means that the first part of the ordered pair is injective, thus the whole thing is injective. Because $A$ will be completely consumed.
      • $(x,g(x))=(y,g(y))$
  • Example:
    • $h(x)=(x,\frac{1}{x})$
    • $f(x)=x$ is injective
  • Check that $h:[0,2\pi )\to \mathbb{R}\times \mathbb{R}$ where $\theta \mapsto (\cos \theta ,\sin \theta )$