Algebra of functions

Definition: Function

A function $f:X\to Y$ consists of three pieces of data (i.e., a triple $(f,X,Y)$):

• A set $X$, which is the domain.
• A set $Y$, which is the codomain.
• A rule $f$ which takes any input element $x\in X$ and produces an unambiguous output $f(x)\in Y$.

A well-defined function produces exactly one value in its codomain for every input from its domain; that is, there must be exactly one such $f(x)$ for each input $x$. An ill-defined function has a formula which is ambiguous, incomplete, or self-contradictory.

Related: Image and preimage, Injectivity, surjectivity, and bijectivity,

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Proof strategies with functions §

• Two functions $f$ and $g$ are equal if:
  • $f:X\to Y$ and $g:X\to Y$ have the same domain and codomain.
  • For all $x\in X$, we have $f(x)=g(x)$.
• To prove a function $f:X\to Y$ is injective, or “one-to-one,” assume $f(x_1)=f(x_2)$ for some $x_1,x_2\in X$ and show $x_1=x_2$.
• To prove a function $f:X\to Y$ is surjective, or “onto,” assume an element $y\in Y$ and show there exists some $x\in X$ such that $f(x)=y$.

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Review §

MATH-UN1207

• Show that $f^{-1}(Y)=X$ no matter what $f,X,Y$ are.
• Suppose $f:\mathbb{R}\to \mathbb{R}$ is the function defined by $f(x)=x^2$.
  • Prove the following by establishing double containment.
    • $f(\mathbb{R})=[0,\infty )$
    • $f([-1,1])=[0,1]$
    • $f([-1,-1/2)\cup [0,1/2])=[0,1]$
  • Calculate the following: $f^{-1}(0)$, $f^{-1}(1)$, $f^{-1}(-1)$.
• Find examples of functions $f$ and sets $S,T,A,B$ so that the sets on the right-hand side are strictly larger:
  • $S\subset f^{-1}(f(S))$
  • $f(f^{-1}(T))\subset T$
  • $f(A\cup B)\subset f(A)\cap f(B)$
• Prove each of the set inclusions above and the equality below using definition pushing arguments.
  • $f(A\cup B)=f(A)\cup f(B)$

MATH-GU4041

• Give an example for $f({\bigcap }_{A\in T}A)\subset {\bigcap }_{A\in T}fA$.
• Show that compositions are associative using a diagram.
• Give an example where function composition does not commute.
• If the composition $(f\circ g)(x)=f(g(x))$ is bijective, is $f$ injective or surjective? What about $g$?