Mathematics 243 – Algebraic Structures Problem Set 2 – Selected Solutions September 18, 2006

Section 1.3/

7. a) Let x ∈ A. Since g is onto, for all y ∈ A there exists x ∈ A such that g(x) = y, but then

f(y) = f(g(x))

= (f ∘ g)(x) by def. of composition

= (h ∘ g)(x) since f ∘ g = h ∘ g (hypothesis) = f(h(x)) by def. of composition = f(y)

Since f(y) = h(y) for all y ∈ A, the mappings are equal.

b) For all x ∈ A, we have (f ∘g)(x) = (f ∘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 ∈ A, the mappings are equal.

For the next two problems, we assume

g : A → B  and  f : B → C.

10. We are given that f ∘ g is onto, so for all z ∈ C, there exists some x ∈ A such that

z = (f ∘ g)(x) = f(g(x)).

But then for all z ∈ C, there exists some y ∈ B, namely y = g(x), such that z = f(y). This shows that f is onto.

11. To show that g is one-to-one, start by assuming g(x) = g(x′) for x, x′ ∈ A. Then “plugging both sides into” f yields f(g(x)) = f(g(x′)). But now we are given f ∘ g is one-to-one, so this implies x = x′. Therefore g is also one-to-one.

13. We have, by the definition of the inverse image of a subset,

x ∈ (f ∘ g)^{−1}(S) ⇔ (f ∘ g)(x) ∈ S
⇔ f(g(x)) ∈ S
⇔ g(x) ∈ f^{−1}(S)
⇔ x ∈ g^{−1} (f^{−1}(S)) .

Hence, (f ∘ g)^{−1}(S) = g^{−1} (f^{−1}(S)) as claimed.