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- [1] For this problem, let h: A ® B be a function from a set A to a set B.
[Note: the symbol ® should be a function arrow.]
We will need some definitions.

**Definition 1**: For any subset C of A, define h(C) to be the subset { h(a) : a is an element of C } of B; we call h(C) the*image*of C under h. It is just the set of all values of h that you get by plugging in elements from C. (Thus given a function h from A to B, we also get a function from the set Subsets(A) of subsets of A to the set Subsets(B) of subsets of B, which is also usually denoted h : Subsets(A) ® Subsets(B). Using the same name for two different things is usually verboten because it is too confusing, but this is what's done in this case. It is not usually a problem, because you can tell which h is meant by seeing what is being plugged into it, either an element or a subset.)

**Definition 2**: For any subset D of B, define h^{-1}(D) to be the subset { a in A : h(a) is in D } of A, called the*inverse image*of D under h. Note that this defines a function h^{-1}: Subsets(B) ® Subsets(A) from the set of subsets of B to the set of subsets of A. (The notation h^{-1}for the inverse image function is unfortunate, since the same notation is used for the inverse function, when it exists: not all functions have an inverse under composition, only the bijective ones do. However, for any function you always have the inverse image function. Again you can tell whether h^{-1}refers to the inverse function or the inverse image function by whether what is being plugged in is an element of B or a subset of B.)

- (a) If f:
**R**®**R**is the function f(x) = x^{2}, find f([1,4]) and f^{ -1}([1,4]), where [a,b] denotes the interval a <= x <= b.

- (b) If h: A ® B is surjective, show that h
^{-1}: Subsets(B) ® Subsets(A) is injective.

- (c) If h: A ® B is injective, is it true that
h
^{-1}: Subsets(B) ® Subsets(A) is surjective? Prove it if it is true, or give a counterexample if it is false.

- (d) Given any subsets E and F of A, show that h(EÇF) is a subset of h(E)Çh(F). Give an example to show that h(EÇF) = h(E)Çh(F) can be false. [Note: Ç should be an intersection symbol.]

- (a) If f:
- [2] Translate each of the following expressions into additive notation:
(a) a
^{-2}(b^{-1}c)^{2}; (b) (ab^{2})^{-3}c^{2}= e.

- [3] Let g be an element of a group G. Define functions

l_{g}: G ® G and r_{g}: G ® G by left and right multiplication, respectively; i.e., l_{g}(x) = gx and r_{g}(x) = xg. Show that l_{g}and r_{g}are bijective. [Note: l should be a Greek letter lambda, and r should be a Greek letter rho.] Use the bijectivity to fill in the following group multiplication table. From the filled in table, determine whether or not the group is abelian:

e a b c d e e a b e b c d e c d a b d

- [4] Let G be a group such that if a, b and c belong to G, and if ab = ca, then b must equal c. Show that G is abelian.