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<h1 id="a-solutions-manual-for-topology-by-james-munkres"><a href="README.html">A solutions manual for Topology by James Munkres</a></h1>
<h3 id="functions">2. Functions</h3>
<p><span id="1"></span><strong>1.</strong> Let <span class="math inline">\(f : A\to B\)</span>. Let <span class="math inline">\(A_0\subset A\)</span> and <span class="math inline">\(B_0\subset B\)</span>.<br />
<span class="math inline">\(\quad\)</span>(a) Show that <span class="math inline">\(A_0\subset f^{-1}(f(A_0))\)</span> and that equality holds if <span class="math inline">\(f\)</span> is injective.<br />
<span class="math inline">\(\quad\)</span>(b) Show that <span class="math inline">\(f(f^{-1}(B_0))\subset B_0\)</span> and that equality holds if <span class="math inline">\(f\)</span> is surjective.</p>
<p><strong><em>Proof.</em></strong> <span class="math inline">\(\quad\)</span>(a) <span class="math inline">\(x\in A_0\Rightarrow f(x)\in f(A_0)\)</span> <span class="math inline">\(\Rightarrow\)</span> <span class="math inline">\(x\in f^{-1}(f(A_0))\)</span>. But if <span class="math inline">\(a\notin A_0\)</span>, <span class="math inline">\(b\in A_0\)</span>, and <span class="math inline">\(f(a)=f(b)\)</span>, then <span class="math inline">\(a\in f^{-1}(f(A_0))\)</span>. If <span class="math inline">\(f\)</span> is injective, then <span class="math inline">\(f(a)=f(b)\)</span> implies <span class="math inline">\(a=b\)</span>. <span class="math inline">\(x\in f^{-1}(f(A_0))\)</span> <span class="math inline">\(\Rightarrow\)</span> <span class="math inline">\(f(x)\in f(A_0)\)</span> <span class="math inline">\(\Rightarrow x\in A_0\)</span>.<br />
<span class="math inline">\(\quad\)</span>(b) <span class="math inline">\(b\in f(f^{-1}(B_0))\Leftrightarrow b=f(a)\)</span> for at least one <span class="math inline">\(a\in f^{-1}(B_0)\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(b=f(a)\)</span> for at least one <span class="math inline">\(a\)</span> such that <span class="math inline">\(f(a)\in B_0\)</span>. <span class="math inline">\(b\notin B_0\Rightarrow\)</span> <span class="math inline">\(b\notin f(f^{-1}(B_0))\)</span>. If <span class="math inline">\(f\)</span> is not surjective, then there is <span class="math inline">\(b\in B_0\)</span> such that <span class="math inline">\(f(a)\neq b\)</span> for every <span class="math inline">\(a\)</span>. Otherwise, for all <span class="math inline">\(b\in B_0\)</span>, <span class="math inline">\(b=f(a)\)</span> for at least one <span class="math inline">\(a\)</span>.<span class="math inline">\(\quad\square\)</span></p>
<p><span id="2"></span><strong>2.</strong> Let <span class="math inline">\(f :A\to B\)</span> and let <span class="math inline">\(A_i\subset A\)</span> and <span class="math inline">\(B_i\subset B\)</span> for <span class="math inline">\(i=0\)</span> and <span class="math inline">\(i=1\)</span>. Show that <span class="math inline">\(f^{-1}\)</span> preserves inclusions, unions, intersections, and differences of sets:<br />
<span class="math inline">\(\quad\)</span>(a) <span class="math inline">\(B_0\subset B_1\Rightarrow f^{-1}(B_0)\subset f^{-1}(B_1)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(b) <span class="math inline">\(f^{-1}(B_0\cup B_1) = f^{-1}(B_0)\cup f^{-1}(B_1)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(c) <span class="math inline">\(f^{-1}(B_0\cap B_1) = f^{-1}(B_0)\cap f^{-1}(B_1)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(d) <span class="math inline">\(f^{-1}(B_0 - B_1) = f^{-1}(B_0) - f^{-1}(B_1)\)</span>.<br />
Show that <span class="math inline">\(f\)</span> preserves inclusions and unions only:<br />
<span class="math inline">\(\quad\)</span>(e) <span class="math inline">\(A_0\subset A_1\Rightarrow f(A_0)\subset f(A_1)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(f) <span class="math inline">\(f(A_0\cup A_1)=f(A_0)\cup f(A_1)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(g) <span class="math inline">\(f (A_0\cap A_1)\subset f (A_0)\cap f (A_1)\)</span>; show that equality holds if <span class="math inline">\(f\)</span> is injective.<br />
<span class="math inline">\(\quad\)</span>(h) <span class="math inline">\(f(A_0-A_1)\supset f (A_0)-f (A_1)\)</span>; show that equality holds if <span class="math inline">\(f\)</span> is injective.</p>
<p><strong><em>Proof.</em></strong> <span class="math inline">\(\quad\)</span>(a) <span class="math inline">\(x\in f^{-1}(B_0)\Rightarrow f(x)\in B_0\)</span> <span class="math inline">\(\Rightarrow\)</span> <span class="math inline">\(f(x)\in B_1\)</span> <span class="math inline">\(\Rightarrow\)</span> <span class="math inline">\(x\in f^{-1}(B_1)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(b) <span class="math inline">\(x\in f^{-1}(B_0\cup B_1)\Leftrightarrow f(x)\in B_0\cup B_1\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(f(x)\in B_1\)</span> or <span class="math inline">\(f(x)\in B_2\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in f^{-1}(B_0)\)</span> or <span class="math inline">\(x\in f^{-1}(B_0)\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(f^{-1}(B_0)\cup f^{-1}(B_1)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(c) <span class="math inline">\(x\in f^{-1}(B_0\cap B_1)\Leftrightarrow f(x)\in B_0\cap B_1\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(f(x)\in B_1\)</span> and <span class="math inline">\(f(x)\in B_2\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in f^{-1}(B_0)\)</span> and <span class="math inline">\(x\in f^{-1}(B_0)\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(f^{-1}(B_0)\cap f^{-1}(B_1)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(d) <span class="math inline">\(x\in f^{-1}(B_0-B_1)\Leftrightarrow f(x)\in B_0-B_1\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(f(x)\in B_1\)</span> and <span class="math inline">\(f(x)\notin B_2\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in f^{-1}(B_0)\)</span> and <span class="math inline">\(x\notin f^{-1}(B_0)\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(f^{-1}(B_0)-f^{-1}(B_1)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(e) <span class="math inline">\(y\in f(A_0)\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(y=f(x)\)</span> for at least one <span class="math inline">\(x\in A_0\)</span> <span class="math inline">\(\Rightarrow\)</span> <span class="math inline">\(y=f(x)\)</span> for at least one <span class="math inline">\(x\in A_1\)</span>.<br />
<span class="math inline">\(\quad\)</span>(f) <span class="math inline">\(y\in f(A_0\cup A_1)\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(y=f(x)\)</span> for at least one <span class="math inline">\(x\)</span> such that <span class="math inline">\(x\in A_0\)</span> or <span class="math inline">\(x\in A_1\)</span> <span class="math inline">\(\Leftrightarrow\)</span> (<span class="math inline">\(y=f(x)\)</span> for at least one <span class="math inline">\(x\in A_0\)</span>) or (<span class="math inline">\(y=f(x)\)</span> for at least one <span class="math inline">\(x\in A_1\)</span>) <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in f(A_0)\cup f(A_1)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(g) Since <span class="math inline">\(A_0\cap A_1\subset A_0\)</span> and <span class="math inline">\(A_0\cap A_1\subset A_1\)</span>, by (e) <span class="math inline">\(f(A_0\cap A_1)\subset f(A_0)\)</span> and <span class="math inline">\(f(A_0\cap A_1)\subset f(A_1)\)</span>. Thus <span class="math inline">\(f(A_0\cap A_1)\subset f(A_0)\cap f(A_1)\)</span>. If <span class="math inline">\(f\)</span> is injective, then <span class="math inline">\(f(a)=f(b)\Rightarrow a=b\)</span>. Thus if <span class="math inline">\(f(x)\in f(A_0)\cap f(A_1)\)</span>, then <span class="math inline">\(x\in A_0\cap A_1\)</span>, and so <span class="math inline">\(f(x)\in f(A_0\cap A_1)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(h) <span class="math inline">\(y\in f(A_0)-f(A_1)\Leftrightarrow y=f(a)\)</span> for at least one <span class="math inline">\(a\in A_0\)</span> and <span class="math inline">\(y\neq f(b)\)</span> for every <span class="math inline">\(b\in A_1\)</span> <span class="math inline">\(\Rightarrow\)</span> <span class="math inline">\(y=f(a)\)</span> for at least one <span class="math inline">\(a\in A-B\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(y\in f(A_0-A_1)\)</span>. If <span class="math inline">\(f\)</span> is injective, then similarly to (g), <span class="math inline">\(f(x)\in f(A_0)-f(A_1)\)</span> <span class="math inline">\(\Rightarrow\)</span> <span class="math inline">\(x\in A_0-A_1\)</span> <span class="math inline">\(\Rightarrow\)</span> <span class="math inline">\(f(x)\in f(A_0-A_1)\)</span>.<span class="math inline">\(\quad\square\)</span></p>
<p><span id="3"></span><strong>3.</strong> Show that (b), (c), (f), and (g) of Exercise 2 hold for arbitrary unions and intersections.</p>
<p><em>Similarly to Exercise 9 of “1. Fundamental Concepts”.</em></p>
<p><span id="4"></span><strong>4.</strong> Let <span class="math inline">\(f:A\to B\)</span> and <span class="math inline">\(g:B\to C\)</span>.<br />
<span class="math inline">\(\quad\)</span>(a) If <span class="math inline">\(C_0\subset C\)</span>, show that <span class="math inline">\((g\circ f)^{-1}(C_0)= f^{-1}(g^{-1} (C_0))\)</span>.<br />
<span class="math inline">\(\quad\)</span>(b) If <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span> are injective, show that <span class="math inline">\(g \circ f\)</span> is injective.<br />
<span class="math inline">\(\quad\)</span>(c) If <span class="math inline">\(g \circ f\)</span> is injective, what can you say about injectivity of <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span>?<br />
<span class="math inline">\(\quad\)</span>(d) If <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span> are surjective, show that <span class="math inline">\(g \circ f\)</span> is surjective.<br />
<span class="math inline">\(\quad\)</span>(e) If <span class="math inline">\(g \circ f\)</span> is surjective, what can you say about surjectivity of <span class="math inline">\(f\)</span> and <span class="math inline">\(g\)</span>?<br />
<span class="math inline">\(\quad\)</span>(f) Summarize your answers to (b)–(e) in the form of a theorem.</p>
<p><strong><em>Proof.</em></strong> <span class="math inline">\(\quad\)</span>(a) <span class="math inline">\(x\in(g\circ f)^{-1}(C_0)\Leftrightarrow\)</span> <span class="math inline">\(g(f(x))\in C_0\Leftrightarrow\)</span> <span class="math inline">\(f(x) \in g^{-1}(C_0)\Leftrightarrow\)</span> <span class="math inline">\(x\in f^{-1}(g^{-1}(C_0))\)</span>.<br />
<span class="math inline">\(\quad\)</span>(b) <span class="math inline">\(g(f(a))\neq g(f(b))\Rightarrow f(a)\neq f(b)\)</span> <span class="math inline">\(\Rightarrow a\neq b\)</span>.<br />
<span class="math inline">\(\quad\)</span>(c) <span class="math inline">\(f\)</span> is injective, but <span class="math inline">\(g\)</span> is not necessarily injective.<br />
<span class="math inline">\(\quad\)</span>(d) <span class="math inline">\((g \circ f)(A)=C\)</span>.<br />
<span class="math inline">\(\quad\)</span>(e) <span class="math inline">\(g\)</span> is surjective, but <span class="math inline">\(f\)</span> is not necessarily surjective.<br />
<span class="math inline">\(\quad\)</span>(f) If <span class="math inline">\(g\circ f\)</span> is bijective, then <span class="math inline">\(f\)</span> is injective, and <span class="math inline">\(g\)</span> is surjective.<span class="math inline">\(\quad\square\)</span></p>
<p><span id="5"></span><strong>5.</strong> In general, let us denote the <em><strong>identity function</strong></em> for a set <span class="math inline">\(C\)</span> by <span class="math inline">\(i_C\)</span>. That is, define <span class="math inline">\(i_C:C\to C\)</span> to be the function given by the rule <span class="math inline">\(i_C(x)=x\)</span> for all <span class="math inline">\(x\in C\)</span>. Given <span class="math inline">\(f: A\to B\)</span>, we say that a function <span class="math inline">\(g : B\to A\)</span> is a <em><strong>left inverse</strong></em> for <span class="math inline">\(f\)</span> if <span class="math inline">\(g \circ f = i_A\)</span>; and we say that <span class="math inline">\(h : B\to A\)</span> is a <em><strong>right inverse</strong></em> for <span class="math inline">\(f\)</span> if <span class="math inline">\(f \circ h = i_B\)</span>.<br />
<span class="math inline">\(\quad\)</span>(a) Show that if <span class="math inline">\(f\)</span> has a left inverse, <span class="math inline">\(f\)</span> is injective; and if <span class="math inline">\(f\)</span> has a right inverse, <span class="math inline">\(f\)</span> is surjective.<br />
<span class="math inline">\(\quad\)</span>(b) Give an example of a function that has a left inverse but no right inverse.<br />
<span class="math inline">\(\quad\)</span>(c) Give an example of a function that has a right inverse but no left inverse.<br />
<span class="math inline">\(\quad\)</span>(d) Can a function have more than one left inverse? More than one right inverse?<br />
<span class="math inline">\(\quad\)</span>(e) Show that if <span class="math inline">\(f\)</span> has both a left inverse <span class="math inline">\(g\)</span> and a right inverse <span class="math inline">\(h\)</span>, then <span class="math inline">\(f\)</span> is bijective and <span class="math inline">\(g=h= f^{-1}\)</span>.</p>
<p><strong><em>Proof.</em></strong> <span class="math inline">\(\quad\)</span>(a) By Exercise 4 (f), <span class="math inline">\(g\circ f\)</span> is bijective <span class="math inline">\(\Rightarrow\)</span> <span class="math inline">\(f\)</span> is injective; <span class="math inline">\(f\circ g\)</span> is bijective <span class="math inline">\(\Rightarrow\)</span> <span class="math inline">\(f\)</span> is surjective.<br />
<span class="math inline">\(\quad\)</span>(b) <span class="math inline">\(f:\{0\}\to\{0,1\}\)</span> given by <span class="math inline">\(x\mapsto x\)</span>.<br />
<span class="math inline">\(\quad\)</span>(c) <span class="math inline">\(f:\{0,1\}\to\{0\}\)</span> given by <span class="math inline">\(x\mapsto 0\)</span>.<br />
<span class="math inline">\(\quad\)</span>(d) Yes.<br />
<span class="math inline">\(\quad\)</span>(e) By Exercise 4 (f), <span class="math inline">\(f\)</span> is bijective. If <span class="math inline">\(h\neq f^{-1}\)</span> or <span class="math inline">\(h\neq f^{-1}\)</span>, then <span class="math inline">\(g\circ f\neq i_A\)</span> or <span class="math inline">\(f\circ h\neq i_B\)</span>. Thus <span class="math inline">\(g=h= f^{-1}\)</span>.<span class="math inline">\(\quad\square\)</span></p>
<p><span id="6"></span><strong>6.</strong> Let <span class="math inline">\(f : \mathbb{R}\to \mathbb{R}\)</span> be the function <span class="math inline">\(f(x) = x^3 - x\)</span>. By restricting the domain and range of <span class="math inline">\(f\)</span> appropriately, obtain from <span class="math inline">\(f\)</span> a bijective function <span class="math inline">\(g\)</span>. Draw the graphs of <span class="math inline">\(g\)</span> and <span class="math inline">\(g^{-1}\)</span>. (There are several possible choices for <span class="math inline">\(g\)</span>.)</p>
<p><strong><em>Solution.</em></strong> <span class="math inline">\(\quad\)</span><span class="math inline">\(g:\{a\}\to\{a^3-a\}\)</span> for every <span class="math inline">\(a\in\mathbb{R}\)</span>, <span class="math inline">\(g:(1,2)\to(0,6)\)</span>, <span class="math inline">\(g:(3,\infty)\to(24,\infty)\)</span>, <span class="math inline">\(\ldots\)</span>.</p>
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