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\centerline{Mathematics 243 -- Algebraic Structures}
\centerline{Problem Set 2 -- Selected Solutions}
\centerline{September 18, 2006}
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\noindent
Section 1.3/
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7. a)  Let $x \in A$.  Since $g$ is onto, for all $y \in A$ there
exists $x \in A$ such that $g(x) = y$, but then 
$$\eqalign{
f(y) &= f(g(x)) \cr
     &= (f\circ g)(x)\quad {\rm by\ def.\ of\ composition}\cr
     &= (h\circ g)(x)\quad {\rm since\ } f\circ g = h\circ g\quad {\rm (hypothesis)}\cr
     &= f(h(x)) \quad {\rm by\ def.\ of\ composition}\cr
     &= f(y)\cr}$$
Since $f(y) = h(y)$ for all $y \in A$, the mappings are equal.
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\noindent
b)  For all $x \in A$, we have $(f\circ g)(x) = (f\circ h)(x)$ by hypothesis.
So $f(g(x)) = f(h(x))$.  Since $f$ is one-to-one, this implies $g(x) = h(x)$.
Since this is true for all $x \in A$, the mappings are equal.
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For the next two problems, we assume 
$$ g: A \to B \qquad {\rm and} \qquad f : B \to C.$$
10.   We are given that $f\circ g$ is onto,
so for all $z \in C$, there exists some $x \in A$ such that 
$$z = (f \circ g)(x) = f(g(x)).$$
But then for all $z \in C$,
there exists some $y \in B$, namely $y = g(x)$, such that $z = f(y)$.
This shows that $f$ is onto.
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\noindent
11.  To show that $g$ is one-to-one, start by assuming 
$g(x) = g(x')$ for $x,x'\in A$.  Then ``plugging
both sides into'' $f$ yields $f(g(x)) = f(g(x'))$.
But now we are given $f\circ g$ is one-to-one,
so this implies $x = x'$.  Therefore $g$ is also one-to-one.
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13.  We have, by the definition of the inverse image of a subset,
$$\eqalign{x \in (f\circ g)^{-1}(S) &\Leftrightarrow (f\circ g)(x) \in S\cr
                                    &\Leftrightarrow f(g(x)) \in S\cr 
                                    &\Leftrightarrow g(x) \in f^{-1}(S)\cr
                                    &\Leftrightarrow x \in g^{-1}\left(f^{-1}(S)\right).\cr}$$
Hence, $(f\circ g)^{-1}(S) = g^{-1}\left(f^{-1}(S)\right)$ as claimed.
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