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Chapter 5: Problem 6

Suppose \(f: A \rightarrow B\) and \(g: B \rightarrow C .\) Hence, \((g \circ f): A \rightarrow C\) exists. Prove (a) If \(f\) and \(g\) are one-to-one, then \(g \circ f\) is one-to-one. (b) If \(f\) and \(g\) are onto mappings, then \(g \circ f\) is an onto mapping. (c) If \(g \circ f\) is one-to-one, then \(f\) is one-to-one. (d) If \(g \circ f\) is an onto mapping, then \(g\) is an onto mapping.

Short Answer

Expert verified
We proved the following statements: (a) If \(f\) and \(g\) are one-to-one, then \(g \circ f\) is one-to-one as well. (b) If \(f\) and \(g\) are onto mappings, then \(g \circ f\) is an onto mapping. (c) If \(g \circ f\) is one-to-one, then \(f\) is one-to-one. (d) If \(g \circ f\) is an onto mapping, then \(g\) is an onto mapping.

Step by step solution

01

a) Proving \((g \circ f)\) is one-to-one when both \(f\) and \(g\) are one-to-one

Given that \(f\) and \(g\) are one-to-one, we must show that \(g \circ f\) is also one-to-one. To prove this, we'll start by assuming two arbitrary elements \(x, y\) in \(A\) such that \((g \circ f)(x) = (g \circ f)(y)\) and then prove that these elements are equal, i.e., \(x = y\). So, let's assume \((g \circ f)(x) = (g \circ f)(y)\), which can be written as \(g(f(x)) = g(f(y))\). Since we know that \(g\) is one-to-one, we can then conclude \(f(x) = f(y)\). Now, using the one-to-one property of \(f\), we can conclude \(x = y\). Therefore, \(g \circ f\) is one-to-one.
02

b) Proving \((g \circ f)\) is an onto mapping when both \(f\) and \(g\) are onto mappings

Given that \(f\) and \(g\) are onto mappings, we must show that \(g \circ f\) is an onto mapping, too. To prove this, we need to show that for every element \(c\) in \(C\), there exists an element \(a\) in \(A\) such that \((g \circ f)(a) = c\). Let \(c \in C\). Since \(g\) is an onto mapping, there must exist an element \(b\) in \(B\) such that \(g(b) = c\). Similarly, as \(f\) is an onto mapping, there must exist an element \(a\) in \(A\) such that \(f(a) = b\). Therefore, \((g \circ f)(a) = g(f(a)) = g(b) = c\). Thus, \(g \circ f\) is an onto mapping.
03

c) Proving \(f\) is one-to-one when \((g \circ f)\) is one-to-one

To prove that \(f\) is one-to-one when \(g \circ f\) is one-to-one, let's assume \(f(x) = f(y)\) for any \(x, y \in A\). We need to show that this implies \(x = y\). Applying \(g\) to both sides of the equation, we get \(g(f(x)) = g(f(y))\). This can be rewritten as \((g \circ f)(x) = (g \circ f)(y)\). Since we know that \((g \circ f)\) is one-to-one, this implies that \(x = y\). Thus, \(f\) is one-to-one.
04

d) Proving \(g\) is an onto mapping when \((g \circ f)\) is an onto mapping

To prove that \(g\) is an onto mapping when \(g \circ f\) is an onto mapping, let's consider any element \(c \in C\). Since \(g\circ f\) is onto, there must exist an element \(a\in A\) such that \((g\circ f)(a) = c\), i.e. \(g(f(a)) = c\). This value \(f(a)\) is an element in \(B\). Let's denote it as \(b\), i.e., \(b = f(a)\). Thus, we have \(g(b) = c\), implying \(g\) is an onto mapping as \(c \in C\) was chosen arbitrarily and a \(b \in B\) exists to match \(c\).

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Most popular questions from this chapter

A diaper liner is placed in each diaper worn by a baby. If, after a diaper change, the liner is soiled, then it is replaced by a new liner. Otherwise, the liner is washed with the diapers and reused, except that each liner is replaced by a new liner after its second use, even if it has never been soiled. The probability that the baby will soil any diaper liner is one-third. If there are only new diaper liners at first, eventually what proportions of the diaper liners being used will be new, once-used, and twice-used?

For each of the following linear operators \(T\) on a vector space \(V\) and ordered bases \(\beta\), compute \([T]_{\beta}\), and determine whether \(\beta\) is a basis consisting of eigenvectors of \(T\). (a) $\mathrm{V}=\mathrm{R}^{2}, \mathrm{~T}\left(\begin{array}{l}a \\\ b\end{array}\right)=\left(\begin{array}{l}10 a-6 b \\ 17 a-10 b\end{array}\right)\(, and \)\beta=\left\\{\left(\begin{array}{l}1 \\\ 2\end{array}\right),\left(\begin{array}{l}2 \\ 3\end{array}\right)\right\\}$ (b) $\quad \mathrm{V}=\mathrm{P}_{1}(R), \mathrm{T}(a+b x)=(6 a-6 b)+(12 a-11 b) x$, and $$ \beta=\\{3+4 x, 2+3 x\\} $$ (c) $\mathrm{V}=\mathrm{R}^{3}, \mathrm{~T}\left(\begin{array}{l}a \\ b \\\ c\end{array}\right)=\left(\begin{array}{r}3 a+2 b-2 c \\ -4 a-3 b+2 c \\\ -c\end{array}\right)$, and $$ \beta=\left\\{\left(\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right),\left(\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right),\left(\begin{array}{l} 1 \\ 0 \\ 2 \end{array}\right)\right\\} $$ (d) \(\mathrm{V}=\mathrm{P}_{2}(R), \mathrm{T}\left(a+b x+c x^{2}\right)=\) $$ (-4 a+2 b-2 c)-(7 a+3 b+7 c) x+(7 a+b+5 c) x^{2}, $$ and \(\beta=\left\\{x-x^{2},-1+x^{2},-1-x+x^{2}\right\\}\) (e) $\mathrm{V}=\mathrm{P}_{3}(R), \mathrm{T}\left(a+b x+c x^{2}+d x^{3}\right)=$ $$ -d+(-c+d) x+(a+b-2 c) x^{2}+(-b+c-2 d) x^{3}, $$ and \(\beta=\left\\{1-x+x^{3}, 1+x^{2}, 1, x+x^{2}\right\\}\) (f) $\mathrm{V}=\mathrm{M}_{2 \times 2}(R), \mathrm{T}\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=\left(\begin{array}{ll}-7 a-4 b+4 c-4 d & b \\\ -8 a-4 b+5 c-4 d & d\end{array}\right)$, and $$ \beta=\left\\{\left(\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right),\left(\begin{array}{rr} -1 & 2 \\ 0 & 0 \end{array}\right),\left(\begin{array}{ll} 1 & 0 \\ 2 & 0 \end{array}\right),\left(\begin{array}{rr} -1 & 0 \\ 0 & 2 \end{array}\right)\right\\} $$

Show that the following mappings \(F, G, H\) are linearly independent: (a) \(F, G, H \in \operatorname{Hom}\left(\mathbf{R}^{2}, \mathbf{R}^{2}\right)\) defined by \(F(x, y)=(x, 2 y), \quad G(x, y)=(y, x+y), \quad H(x, y)=(0, x)\) (b) \(\quad F, G, H \in \operatorname{Hom}\left(\mathbf{R}^{3}, \mathbf{R}\right)\) defined by \(F(x, y, z)=x+y+z, \quad G(x, y, z)=y+z, \quad H(x, y, z)=x-z\)

Let \(V\) be of finite dimension and let \(T\) be a linear operator on \(V\) for which \(T R=I\), for some operator \(R\) on \(V\). (We call \(R\) a right inverse of \(T\).) (a) Show that \(T\) is invertible. (b) Show that \(R=T^{-1}\) (c) Give an example showing that the above need not hold if \(V\) is of infinite dimension.

Show that each of the following linear operators \(T\) on \(\mathbf{R}^{3}\) is nonsingular and find a formula for \(T^{-1}\), where (a) \(T(x, y, z)=(x-3 y-2 z, y-4 z, z)\) (b) \(T(x, y, z)=(x+z, x-y, y)\)
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