A solutions manual for Topology by James Munkres
2. Functions
1. Let \(f : A\to B\). Let \(A_0\subset A\) and \(B_0\subset B\).
\(\quad\)(a) Show that \(A_0\subset f^{-1}(f(A_0))\) and that equality holds if \(f\) is injective.
\(\quad\)(b) Show that \(f(f^{-1}(B_0))\subset B_0\) and that equality holds if \(f\) is surjective.
Proof. \(\quad\)(a) \(x\in A_0\Rightarrow f(x)\in f(A_0)\) \(\Rightarrow\) \(x\in f^{-1}(f(A_0))\). But if \(a\notin A_0\), \(b\in A_0\), and \(f(a)=f(b)\), then \(a\in f^{-1}(f(A_0))\). If \(f\) is injective, then \(f(a)=f(b)\) implies \(a=b\). \(x\in f^{-1}(f(A_0))\) \(\Rightarrow\) \(f(x)\in f(A_0)\) \(\Rightarrow x\in A_0\).
\(\quad\)(b) \(b\in f(f^{-1}(B_0))\Leftrightarrow b=f(a)\) for at least one \(a\in f^{-1}(B_0)\) \(\Leftrightarrow\) \(b=f(a)\) for at least one \(a\) such that \(f(a)\in B_0\). \(b\notin B_0\Rightarrow\) \(b\notin f(f^{-1}(B_0))\). If \(f\) is not surjective, then there is \(b\in B_0\) such that \(f(a)\neq b\) for every \(a\). Otherwise, for all \(b\in B_0\), \(b=f(a)\) for at least one \(a\).\(\quad\square\)
2. Let \(f :A\to B\) and let \(A_i\subset A\) and \(B_i\subset B\) for \(i=0\) and \(i=1\). Show that \(f^{-1}\) preserves inclusions, unions, intersections, and differences of sets:
\(\quad\)(a) \(B_0\subset B_1\Rightarrow f^{-1}(B_0)\subset f^{-1}(B_1)\).
\(\quad\)(b) \(f^{-1}(B_0\cup B_1) = f^{-1}(B_0)\cup f^{-1}(B_1)\).
\(\quad\)(c) \(f^{-1}(B_0\cap B_1) = f^{-1}(B_0)\cap f^{-1}(B_1)\).
\(\quad\)(d) \(f^{-1}(B_0 - B_1) = f^{-1}(B_0) - f^{-1}(B_1)\).
Show that \(f\) preserves inclusions and unions only:
\(\quad\)(e) \(A_0\subset A_1\Rightarrow f(A_0)\subset f(A_1)\).
\(\quad\)(f) \(f(A_0\cup A_1)=f(A_0)\cup f(A_1)\).
\(\quad\)(g) \(f (A_0\cap A_1)\subset f (A_0)\cap f (A_1)\); show that equality holds if \(f\) is injective.
\(\quad\)(h) \(f(A_0-A_1)\supset f (A_0)-f (A_1)\); show that equality holds if \(f\) is injective.
Proof. \(\quad\)(a) \(x\in f^{-1}(B_0)\Rightarrow f(x)\in B_0\) \(\Rightarrow\) \(f(x)\in B_1\) \(\Rightarrow\) \(x\in f^{-1}(B_1)\).
\(\quad\)(b) \(x\in f^{-1}(B_0\cup B_1)\Leftrightarrow f(x)\in B_0\cup B_1\) \(\Leftrightarrow\) \(f(x)\in B_1\) or \(f(x)\in B_2\) \(\Leftrightarrow\) \(x\in f^{-1}(B_0)\) or \(x\in f^{-1}(B_0)\) \(\Leftrightarrow\) \(f^{-1}(B_0)\cup f^{-1}(B_1)\).
\(\quad\)(c) \(x\in f^{-1}(B_0\cap B_1)\Leftrightarrow f(x)\in B_0\cap B_1\) \(\Leftrightarrow\) \(f(x)\in B_1\) and \(f(x)\in B_2\) \(\Leftrightarrow\) \(x\in f^{-1}(B_0)\) and \(x\in f^{-1}(B_0)\) \(\Leftrightarrow\) \(f^{-1}(B_0)\cap f^{-1}(B_1)\).
\(\quad\)(d) \(x\in f^{-1}(B_0-B_1)\Leftrightarrow f(x)\in B_0-B_1\) \(\Leftrightarrow\) \(f(x)\in B_1\) and \(f(x)\notin B_2\) \(\Leftrightarrow\) \(x\in f^{-1}(B_0)\) and \(x\notin f^{-1}(B_0)\) \(\Leftrightarrow\) \(f^{-1}(B_0)-f^{-1}(B_1)\).
\(\quad\)(e) \(y\in f(A_0)\) \(\Leftrightarrow\) \(y=f(x)\) for at least one \(x\in A_0\) \(\Rightarrow\) \(y=f(x)\) for at least one \(x\in A_1\).
\(\quad\)(f) \(y\in f(A_0\cup A_1)\) \(\Leftrightarrow\) \(y=f(x)\) for at least one \(x\) such that \(x\in A_0\) or \(x\in A_1\) \(\Leftrightarrow\) (\(y=f(x)\) for at least one \(x\in A_0\)) or (\(y=f(x)\) for at least one \(x\in A_1\)) \(\Leftrightarrow\) \(x\in f(A_0)\cup f(A_1)\).
\(\quad\)(g) Since \(A_0\cap A_1\subset A_0\) and \(A_0\cap A_1\subset A_1\), by (e) \(f(A_0\cap A_1)\subset f(A_0)\) and \(f(A_0\cap A_1)\subset f(A_1)\). Thus \(f(A_0\cap A_1)\subset f(A_0)\cap f(A_1)\). If \(f\) is injective, then \(f(a)=f(b)\Rightarrow a=b\). Thus if \(f(x)\in f(A_0)\cap f(A_1)\), then \(x\in A_0\cap A_1\), and so \(f(x)\in f(A_0\cap A_1)\).
\(\quad\)(h) \(y\in f(A_0)-f(A_1)\Leftrightarrow y=f(a)\) for at least one \(a\in A_0\) and \(y\neq f(b)\) for every \(b\in A_1\) \(\Rightarrow\) \(y=f(a)\) for at least one \(a\in A-B\) \(\Leftrightarrow\) \(y\in f(A_0-A_1)\). If \(f\) is injective, then similarly to (g), \(f(x)\in f(A_0)-f(A_1)\) \(\Rightarrow\) \(x\in A_0-A_1\) \(\Rightarrow\) \(f(x)\in f(A_0-A_1)\).\(\quad\square\)
3. Show that (b), (c), (f), and (g) of Exercise 2 hold for arbitrary unions and intersections.
Similarly to Exercise 9 of “1. Fundamental Concepts”.
4. Let \(f:A\to B\) and \(g:B\to C\).
\(\quad\)(a) If \(C_0\subset C\), show that \((g\circ f)^{-1}(C_0)= f^{-1}(g^{-1} (C_0))\).
\(\quad\)(b) If \(f\) and \(g\) are injective, show that \(g \circ f\) is injective.
\(\quad\)(c) If \(g \circ f\) is injective, what can you say about injectivity of \(f\) and \(g\)?
\(\quad\)(d) If \(f\) and \(g\) are surjective, show that \(g \circ f\) is surjective.
\(\quad\)(e) If \(g \circ f\) is surjective, what can you say about surjectivity of \(f\) and \(g\)?
\(\quad\)(f) Summarize your answers to (b)–(e) in the form of a theorem.
Proof. \(\quad\)(a) \(x\in(g\circ f)^{-1}(C_0)\Leftrightarrow\) \(g(f(x))\in C_0\Leftrightarrow\) \(f(x) \in g^{-1}(C_0)\Leftrightarrow\) \(x\in f^{-1}(g^{-1}(C_0))\).
\(\quad\)(b) \(g(f(a))\neq g(f(b))\Rightarrow f(a)\neq f(b)\) \(\Rightarrow a\neq b\).
\(\quad\)(c) \(f\) is injective, but \(g\) is not necessarily injective.
\(\quad\)(d) \((g \circ f)(A)=C\).
\(\quad\)(e) \(g\) is surjective, but \(f\) is not necessarily surjective.
\(\quad\)(f) If \(g\circ f\) is bijective, then \(f\) is injective, and \(g\) is surjective.\(\quad\square\)
5. In general, let us denote the identity function for a set \(C\) by \(i_C\). That is, define \(i_C:C\to C\) to be the function given by the rule \(i_C(x)=x\) for all \(x\in C\). Given \(f: A\to B\), we say that a function \(g : B\to A\) is a left inverse for \(f\) if \(g \circ f = i_A\); and we say that \(h : B\to A\) is a right inverse for \(f\) if \(f \circ h = i_B\).
\(\quad\)(a) Show that if \(f\) has a left inverse, \(f\) is injective; and if \(f\) has a right inverse, \(f\) is surjective.
\(\quad\)(b) Give an example of a function that has a left inverse but no right inverse.
\(\quad\)(c) Give an example of a function that has a right inverse but no left inverse.
\(\quad\)(d) Can a function have more than one left inverse? More than one right inverse?
\(\quad\)(e) Show that if \(f\) has both a left inverse \(g\) and a right inverse \(h\), then \(f\) is bijective and \(g=h= f^{-1}\).
Proof. \(\quad\)(a) By Exercise 4 (f), \(g\circ f\) is bijective \(\Rightarrow\) \(f\) is injective; \(f\circ g\) is bijective \(\Rightarrow\) \(f\) is surjective.
\(\quad\)(b) \(f:\{0\}\to\{0,1\}\) given by \(x\mapsto x\).
\(\quad\)(c) \(f:\{0,1\}\to\{0\}\) given by \(x\mapsto 0\).
\(\quad\)(d) Yes.
\(\quad\)(e) By Exercise 4 (f), \(f\) is bijective. If \(h\neq f^{-1}\) or \(h\neq f^{-1}\), then \(g\circ f\neq i_A\) or \(f\circ h\neq i_B\). Thus \(g=h= f^{-1}\).\(\quad\square\)
6. Let \(f : \mathbb{R}\to \mathbb{R}\) be the function \(f(x) = x^3 - x\). By restricting the domain and range of \(f\) appropriately, obtain from \(f\) a bijective function \(g\). Draw the graphs of \(g\) and \(g^{-1}\). (There are several possible choices for \(g\).)
Solution. \(\quad\)\(g:\{a\}\to\{a^3-a\}\) for every \(a\in\mathbb{R}\), \(g:(1,2)\to(0,6)\), \(g:(3,\infty)\to(24,\infty)\), \(\ldots\).