Suppose "g : A\\to B" and "f : B\\to C" are functions.
a. Show that if "f\\circ g" is onto, then "f" must also be onto.
Answer. Assume that "f\\circ g" is onto. Let "c\\in C" . Since "f\\circ g" is onto, there is "a\\in A" such that "c = (f\\circ g)(a)" by the definition of onto function. By the definition of function composition, "c = f(g(a))" . Let "b = g(a)" . This "b" belongs to "B" and is such that "c = f(b)". Therefore, for every "c\\in C" , there is "b\\in B" such that "c = f(b)" . By the definition of onto function, "f" is onto.
b. Show that if "f\\circ g" is one-to-one, then "g" must also be one-to-one.
Answer. Let "a_1, a_2\\in A" such that "g(a_1) = g(a_2)" . Then
"(f\\circ g)(a_1) = f(g(a_1)) = f(g(a_2))= (f\\circ g)(a_2)."Since "f\\circ g" is one-to-one, "a_1=a_2" . Therefore, for all "a_1, a_2\\in A" such that "g(a_1) = g(a_2)" , "a_1=a_2" . By the definition of one-to-one function, "g" is one-to-one.
c. Show that if "f\\circ g" is a bijection, then "g" is onto if and only if "f" is one-to-one.
Answer. Assume that "f\\circ g" is a bijection.
Assume that "g" is onto. Let "b_1, b_2\\in B" such that "f(b_1) = f(b_2)" . Since "g" is onto, by the definition of onto function, there are "a_1, a_2\\in A" such that "b_1 = g(a_1)" and "b_2 = g(a_2)" . We have
"(f\\circ g)(a_1) = f(g(a_1)) = f(b_1)""= f(b_2) = f(g(a_2)) = (f\\circ g)(a_2)."
Since "f\\circ g" is a bijection, it is one-to-one, hence "a_1=a_2" , and
"b_1 = g(a_1) = g(a_2) = b_2."
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