Let \(f: A \to B\) and \(g: B \to C\) .
If \(C_0 \subset C\), show that \((g \circ f)^{-1}(C_0) = f^{-1}(g^{-1}(C_0))\) .
If \(f\) and \(g\) are injective, show that \(g \circ f\) is injective.
If \(g \circ f\) is injective, what can you say about injectivity of \(f\) and \(g\) ?
If \(f\) and \(g\) are surjective, show that \(g \circ f\) is surjective.
If \(g \circ f\) is surjective, what can you say about surjectivity of \(f\) and \(g\) ?
Summarize your answers to (b)-(e) in the form of a theorem.
Solution
For \(a\in A\), \[ a\in(g\circ f)^{-1}(C_0) \iff g(f(a))\in C_0 \iff f(a)\in g^{-1}(C_0) \iff a\in f^{-1}(g^{-1}(C_0)). \]
If \(g(f(a_1))=g(f(a_2))\), injectivity of \(g\) gives \(f(a_1)=f(a_2)\), and injectivity of \(f\) then gives \(a_1=a_2\). Thus \(g\circ f\) is injective.
If \(g\circ f\) is injective, then \(f\) must be injective: \(f(a_1)=f(a_2)\) implies \(g(f(a_1))=g(f(a_2))\). Nothing follows about injectivity of \(g\) away from \(f(A)\). For example, the inclusion \(f:\{0\}\to\{0,1\}\) followed by the constant map \(g:\{0,1\}\to\{*\}\) has injective composite from a singleton, while \(g\) is not injective.
Given \(c\in C\), choose \(b\in B\) with \(g(b)=c\), then \(a\in A\) with \(f(a)=b\). Thus \((g\circ f)(a)=c\), so the composite is surjective.
If \(g\circ f\) is surjective, then \(g\) is surjective: every \(c\in C\) equals \(g(f(a))\) for some \(a\). Nothing follows about surjectivity of \(f\). For example, the inclusion \(f:\{0\}\to\{0,1\}\) followed by the constant map \(g:\{0,1\}\to\{*\}\) has surjective composite although \(f\) is not surjective.
Therefore:
- the composite of injective maps is injective, and injectivity of a composite forces the first map to be injective;
- the composite of surjective maps is surjective, and surjectivity of a composite forces the second map to be surjective.