Topology: An Inquiry-Based Approach

Find a function $f:\mathbb{R}\to \mathbb{R}$ such that each element in the codomain has exactly one preimage.

Find a function $f:\mathbb{R}\to \mathbb{R}$ such that each element in the codomain has at least two preimages.

Find a function $f:\mathbb{R}\to \mathbb{R}$ such that each element has exactly two preimages.

Find a function $f:\mathbb{R}\to \mathbb{R}$ such that there is an element in the codomain that has exactly three preimages and another element in the codomain that as exactly two preminages.

Find a function $f:\mathbb{R}\to \mathbb{R}$ such that there is an element in the codomain that has infinitely many preimages.

$F:\mathbb{R}\to \mathbb{R}$ defined by $F(x)=5x+3\text{,}$ for all $x\in \mathbb{R}$

$G:\mathbb{Z}\to \mathbb{Z}$ defined by $G(x)=5x+3\text{,}$ for all $x\in \mathbb{Z}$

$f:(\mathbb{R}\setminus \left\{4\right\})\to \mathbb{R}$ defined by $f(x)={\displaystyle \frac{3x}{x-4}}\text{,}$ for all $x\in (\mathbb{R}\setminus \{4\})$

$g:(\mathbb{R}\setminus \left\{4\right\})\to (\mathbb{R}\setminus \left\{3\right\})$ defined by $g(x)={\displaystyle \frac{3x}{x-4}}\text{,}$ for all $x\in (\mathbb{R}\setminus \{4\})$

$h:\mathbb{R}\to {\mathbb{R}}_{\ge 0}$ defined by $h(x)=x^2$ for every $x\in \mathbb{R}\text{,}$ where ${\mathbb{R}}_{\ge 0}=\{x\in \mathbb{R}\mid x\ge 0\}$

$k:{\mathbb{R}}_{\ge 0}\to {\mathbb{R}}_{\ge 0}$ defined by $k(x)=x^2$ for every $x\in {\mathbb{R}}_{\ge 0}$

If $f$ an injection? Is $f$ a surjection? Explain.

Find a largest subset $C$ of $A$ (largest in the number of elements of $C$) such that $f{|}_C$ is an injection.

Find a subset $D$ of $B$ such that $f$ is a surjection.

Find subsets $X$ of $A$ and $Y$ of $B$ such that $f{|}_X:X\to Y$ is a bijection.

Illustrate a subset $X\subset A\times B$ that is the Cartesian product of a subset of $A$ with a subset of $B\text{.}$

Show that there is a subset $W\subset A\times B$ that is not the Cartesian product of a subset of $A$ with a subset of $B\text{.}$ [Thus, not every subset of a Cartesian product is the Cartesian product of a pair of subsets.]

Let $A$ be a subset of $X\text{.}$ Show that $A\subseteq f^{-1}(f(A))\text{.}$ Make an example to show that in general, $A\ne f^{-1}(f(A))\text{.}$

Let $B$ be a subset of $Y\text{.}$ Show that $f(f^{-1}(B))\subseteq B\text{.}$ Make an example to show that in general, $f(f^{-1}(B))\ne B\text{.}$

Prove that $f$ is a surjection if and only if $f(f^{-1}(B))=B$ for every subset $B$ of $Y\text{.}$

Prove that $f$ is an injection if and only if $f^{-1}(f(A))=A$ for every subset $A$ of $X\text{.}$

For every $x$ in $A\text{,}$ $(f^{-1}\circ f)(x)=x\text{.}$

For every $y$ in $B\text{,}$ $(f\circ f^{-1})(y)=y\text{.}$

Prove that ${\pi }_i$ is a surjection for each $i\text{.}$

Prove that there is a unique function $f:X\to Y$ such that ${\pi }_i\circ f=f_i\circ {\pi }_i$ for each $i\text{.}$ (Note that one of the ${\pi }_i$ maps $X$ to $X_i$ and the other maps $Y$ to $Y_i\text{.}$)

Suppose that each $f_i$ has an inverse $h_i\text{.}$ Show that ${(f_1\times f_2)}^{-1}=h_1\times h_1\text{.}$

Is the function $f$ an injection? Justify your conclusion.

Is the function $f$ a surjection? Justify your conclusion.

Is it true that if $g\circ f$ is an injection, then both $f$ and $g$ are injections? If the answer is no, are there any conditions that $f$ or $g$ must satisfy if $g\circ f$ an injection? Prove your answers.

Is it true that if $g\circ f$ is a surjection, then both $f$ and $g$ are surjections? If the answer is no, are there any conditions that $f$ or $g$ must satisfy if $f\circ g$ a surjection? Prove your answers.

Is composition of functions a commutative operation? Prove your answer.

Is composition of functions an associative operation? Prove your answer.

Define $f:{\mathbb{Z}}_5\to {\mathbb{Z}}_5$ by $f([x])=[x^2+4]$ for all $[x]\in {\mathbb{Z}}_5\text{.}$ Write the inverse of $f$ as a set of ordered pairs, and explain why $f^{-1}$ is not a function.

Define $g:{\mathbb{Z}}_5\to {\mathbb{Z}}_5$ by $g([x])=[x^3+4]$ for all $[x]\in {\mathbb{Z}}_5\text{.}$ Write the inverse of $g$ as a set of ordered pairs, and explain why $g^{-1}$ is a function.

Is it possible to write a formula for $g^{-1}([y])\text{,}$ where $[y]\in {\mathbb{Z}}_5\text{?}$ The answer to this question depends on whether or not it is possible to define a cube root of elements of ${\mathbb{Z}}_5\text{.}$ Recall that for a real number $x\text{,}$ we define the cube root of $x$ to be the real number $y$ such that $y^3=x\text{.}$ That is, $y=\sqrt[3]{x}$ if and only if $y^3=x\text{.}$ Using this idea, is it possible to define the cube root of each element of ${\mathbb{Z}}_5\text{?}$ If so, what is $\sqrt[3]{[0]}\text{,}$ $\sqrt[3]{[1]}\text{,}$ $\sqrt[3]{[2]}\text{,}$ $\sqrt[3]{[3]}\text{,}$ and $\sqrt[3]{[4]}\text{.}$

Now answer the question posed at the beginning of part (c). If possible, determine a formula for $g^{-1}([y])$ where $g^{-1}:{\mathbb{Z}}_5\to {\mathbb{Z}}_5\text{.}$

If $A$ is a subset of $X\text{,}$ then $A\subseteq f^{-1}(f(A))\text{.}$

If $A$ is a subset of $X\text{,}$ then $f^{-1}(f(A))\subseteq A\text{.}$

If $B$ is a subset of $Y\text{,}$ then $B\subseteq f(f^{-1}(B))\text{.}$

If $B$ is a subset of $Y\text{,}$ then $f(f^{-1}(B))\subseteq B\text{.}$

If $A_1$ and $A_2$ are subsets of $X$ with $A_1\subseteq A_2\text{,}$ then $f(A_1)\subseteq f(A_2)\text{.}$

If $B_1$ and $B_2$ are subsets of $Y$ with $B_1\subseteq B_2\text{,}$ then $f^{-1}(B_1)\subseteq f^{-1}(B_2)\text{.}$

If $B_1$ and $B_2$ are subsets of $Y$ with $B_1\subseteq B_2\text{,}$ then $f^{-1}(B_2)\subseteq f^{-1}(B_1)\text{.}$

If $A_1$ and $A_2$ are subsets of $X\text{,}$ then $f(A_1\cup A_2)=f(A_1)\cup f(A_2)\text{.}$

If $B_1$ and $B_2$ are subsets of $Y\text{,}$ then $f^{-1}(B_1\cup B_2)=f^{-1}(B_1)\cup f^{-1}(B_2)\text{.}$

If $A_1$ and $A_2$ are subsets of $X\text{,}$ then $f(A_1\cap A_2)=f(A_1)\cap f(A_2)\text{.}$

If $B_1$ and $B_2$ are subsets of $Y\text{,}$ then $f^{-1}(B_1\cap B_2)=f^{-1}(B_1)\cap f^{-1}(B_2)\text{.}$

If $A_1$ and $A_2$ are subsets of $X\text{,}$ then $f(A_1\setminus A_2)=f(A_1)\setminus f(A_2)\text{.}$

If $B_1$ and $B_2$ are subsets of $Y\text{,}$ then $f^{-1}(B_1\setminus B_2)=f^{-1}(B_1)\setminus f^{-1}(B_2)\text{.}$