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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.

f([1,4]) = [1, 16] and f^{ -1}([1,4]) is the union of [-2, -1] and [1,2]

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

Let E and F be subsets of B such that h^{-1}(E) = h^{-1}(F). Let b be any element of E; since h is surjective, there is an a with h(a) = b. By definition, a is in h^{-1}(E). But h^{-1}(E) = h^{-1}(F), so b = h(a) is in F. I.e., every element of E is also in F. Thus F contains E. The same argument, starting with an element of F shows that E contains F. Thus E = F, so h^{-1}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.

It is true. Let D be any subset of A. Let E = h(D). Then h^{-1}(E) is the set of all elements x of A such that h(x) is in E. But since h is injective, the only elements of A that map to E are the elements of D. Thus D = h^{-1}(E), so h^{-1}is surjective.

- (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.]

Let b be an element of h(EÇF). Thus b = h(a) for some element a of EÇF. But this means that a is in both E and F so h(a) is in both h(E) and h(F), hence b=h(a) is in h(E)Çh(F). Thus h(E)Çh(F) contains h(EÇF).

To show that h(EÇF) = h(E)Çh(F) can be false, let h:**R**®**R**be the function h(x) = x^{2}, and let E = [-1,0] and let F = [0,1]. Then EÇF = {0}, so h(EÇF) = {0}, but h(E) = [0,1] = h(F), so h(E)Çh(F) = [0,1].

- (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.

(a) -2a + 2(-b + c), hence, assuming commutativity, -2a - 2b + 2c

(b) -3(a + 2b) + 2c = 0, hence, assuming commutativity, -2a - 2b + 2c -3a - 6b + 2c = 0

- [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:

Note that r_{g}r_{g-1}(x) = (xg^{-1})g = xe = x so r_{g}r_{g-1}= 1_{G}, and r_{g-1}r_{g}(x) = (xg)g^{-1}= xe = x so r_{g-1}r_{g}= 1_{G}. Thus r_{g}is injective and surjective, hence bijective. The proof for l_{g}is the same, except the multiplications are on the left (gx instead of xg).

To fill in the table, we first fill in the row and column corresponding to multiplication by e. We can now fill in the b row (since l_{b}is bijective, we know every element of G must appear in this row, so the one spot not already filled must be filled by the missing element, which is a). Likewise, we can now fill the c row and the a and d columns. At this point we see that the ab entry must be c, since every other element of the group is in either the a row or the b column. Now an a goes in the last open spot of the b column, and we easily fill in the c column because each row is short at most one entry.

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

Because the table is symmetric, we see that xy = yx no matter what x and y are. Thus the group is abelian.

- [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.

We must show that xy = yx for all elements x and y of G. But xyx^{-1}equals something; say xyx^{-1}= z. Then (by multiplying through on the right by x) we have xy = zx, so by hypothesis y = z, hence xy = yx. Thus G is abelian.