Show That a Continuous Map F S 1 R Can Be Neither Injective Nor Surjective

A function $f \colon X\to Y$ is a rule that, for every element $x\in X,$ associates an element $f(x) \in Y.$ The element $f(x)$ is sometimes called the image of $x,$ and the subset of $Y$ consisting of images of elements in $X$ is called the image of $f.$ That is,

$\text{image}(f) = \{ y \in Y : y = f(x) \text{ for some } x \in X\}.$

Let $f \colon X \to Y$ be a function. Then $f$ is injective if distinct elements of $X$ are mapped to distinct elements of $Y.$

That is, if $x_1$ and $x_2$ are in $X$ such that $x_1 \ne x_2$ , then $f(x_1) \ne f(x_2)$ .

This is equivalent to saying if $f(x_1) = f(x_2)$ , then $x_1 = x_2$ .

A synonym for "injective" is "one-to-one."

The function $f\colon {\mathbb Z} \to {\mathbb Z}$ defined by $f(n) = 2n$ is injective: if $2x_1=2x_2,$ dividing both sides by $2$ yields $x_1=x_2.$

The function $f\colon {\mathbb Z} \to {\mathbb Z}$ defined by $f(n) = \big\lfloor \frac n2 \big\rfloor$ is not injective; for example, $f(2) = f(3) = 1$ but $2 \ne 3.$

The function $f\colon \{ \text{German football players dressed for the 2014 World Cup final}\} \to {\mathbb N}$ defined by $f(A) = \text{the jersey number of } A$ is injective; no two players were allowed to wear the same number.

The existence of an injective function gives information about the relative sizes of its domain and range:

If $X$ and $Y$ are finite sets and $f\colon X\to Y$ is injective, then $|X| \le |Y|.$

$x$ $y$ $z$ None of the above

Let $f$ be a one-to-one (Injective) function with domain $D_{f} = \{x,y,z\}$ and range $\{1,2,3\}.$ It is given that only one of the following $3$ statement is true and the remaining statements are false:

$\begin{aligned} f(x) &=& 1 \\ f(y) & \neq & 1 \\ f(z)& \neq & 2. \\ \end{aligned}$

Find $f^{-1} (1).$

Let $f \colon X\to Y$ be a function. Then $f$ is surjective if every element of $Y$ is the image of at least one element of $X.$ That is, $\text{image}(f) = Y.$

Symbolically,

$\forall y \in Y, \exists x \in X \text{ such that } f(x) = y.$

A synonym for "surjective" is "onto."

The function $f\colon {\mathbb Z} \to {\mathbb Z}$ defined by $f(n) = 2n$ is not surjective: there is no integer $n$ such that $f(n)=3,$ because $2n=3$ has no solutions in $\mathbb Z.$ So $3$ is not in the image of $f.$

The function $f\colon {\mathbb Z} \to {\mathbb Z}$ defined by $f(n) = \big\lfloor \frac n2 \big\rfloor$ is surjective. For any integer $m,$ note that $f(2m) = \big\lfloor \frac{2m}2 \big\rfloor = m,$ so $m$ is in the image of $f.$ So the image of $f$ equals $\mathbb Z.$

The function $f \colon \{\text{US senators}\} \to \{\text{US states}\}$ defined by $f(A) = \text{the state that } A \text{ represents}$ is surjective; every state has at least one senator.

The existence of a surjective function gives information about the relative sizes of its domain and range:

If $X$ and $Y$ are finite sets and $f\colon X\to Y$ is surjective, then $|X| \ge |Y|.$

Let $E = \{1, 2, 3, 4\}$ and $F = \{1, 2\}.$ Then what is the number of onto functions from $E$ to $F?$

A function is bijective for two sets if every element of one set is paired with only one element of a second set, and each element of the second set is paired with only one element of the first set. This means that all elements are paired and paired once.

Let $f \colon X \to Y$ be a function. Then $f$ is bijective if it is injective and surjective; that is, every element $y \in Y$ is the image of exactly one element $x \in X.$

The function $f \colon {\mathbb R} \to {\mathbb R}$ defined by $f(x) = 2x$ is a bijection.

The function $f \colon {\mathbb Z} \to {\mathbb Z}$ defined by $f(n) = \begin{cases} n+1 &\text{if } n \text{ is odd} \\ n-1&\text{if } n \text{ is even}\end{cases}$ is a bijection.

The function $f\colon \{ \text{months of the year}\} \to \{1,2,3,4,5,6,7,8,9,10,11,12\}$ defined by $f(M) = \text{ the number } n \text{ such that } M \text{ is the } n^\text{th} \text{ month}$ is a bijection.

Note that the above discussions imply the following fact (see the Bijective Functions wiki for examples):

If $X$ and $Y$ are finite sets and $f\colon X\to Y$ is bijective, then $|X| = |Y|.$

The following alternate characterization of bijections is often useful in proofs:

Suppose $X$ is nonempty. Then $f \colon X \to Y$ is a bijection if and only if there is a function $g\colon Y \to X$ such that $g \circ f$ is the identity on $X$ and $f\circ g$ is the identity on $Y;$ that is, $g\big(f(x)\big)=x$ and $f\big(g(y)\big)=y$ for all $x\in X, y \in Y.$ When this happens, the function $g$ is called the inverse function of $f$ and is also a bijection.

Show that the function $f\colon {\mathbb R} \to {\mathbb R}$ defined by $f(x)=x^3$ is a bijection.

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Rather than showing $f$ is injective and surjective, it is easier to define $g\colon {\mathbb R} \to {\mathbb R}$ by $g(x) = x^{1/3}$ and to show that $g$ is the inverse of $f.$ This follows from the identities $\big(x^3\big)^{1/3} = \big(x^{1/3}\big)^3 = x.$ $\big($ Followup question: the same proof does not work for $f(x) = x^2.$ Why not? $\big)$

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Source: https://brilliant.org/wiki/bijection-injection-and-surjection/

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