Bridge to Abstract Mathematics: Mathematical Proof and Structures

8.2 MORE ON FUNCTIONS AND MAPPINGS 271

(c) A = f-'(B)

(d) rng f = f(A)

(e) If MI, M, c A with M, c M,, then f(M,) G f(M2)

(f) If N,, N2 E B with N, E N2, then f-'(N,) c f-'(N,)

Partial proof (a) To prove f(@) = @, we recall the approach of Examples

11 and 12, Article 4.1, and assume y E f(@). By definition of image, we

may then assert there exists m E @ such that y = f(m). But the statement

"3m E @" is a contradiction.

(f) Let N, and N, be subsets of B with N, E N,. To prove

f - '(N ,) G f - '(N,), suppose x E f - '(N ,). To show x E f - '(N,), we need

only show f (x) E N,. Now since x E f - '(N,), we have f (x) E N ,. Since

N, E N,, we conclude f (x) E N,, as desired.

Let us next consider an example. If f (x) = x2 maps R to R, then
f ([ - 1, 11) = [0, 11, whereas f - '([0,4]) = [ - 2,2]. Also, f - '((9)) =
( - 3,3), whereas f - I(( - 9)) = @. Furthermore, f (R) = [0, a), the range
off.
Let us compute f - (f (0 1)) Now f ([0, 11) = [0, 11, so
f -l(f([O, 11)) = f -'([O, I]) = [- 1, 11. Note that letting M = [0, 11, we
have that f - '( f (M)) = [ - 1,1] 3 [O,1] = M in this example. On the other
hand, f(f -I([- 1, 11)) = f([- 1, 11) = [0, 11. Letting N = [- 1, 11, we
note that f( f -'(N)) = [0, 11 c [- 1, 11 = N in this example.
The results of several parts of the preceding example are suggestive of
part of the next result.

THEOREM 5

Given a mapping f: A -+ B.

(a) M G f-'(f(M)) for any subset M of A

(b) f(f-'(N)) c N for any subset N of B

(c) M = f-'(f(M)) for each subset M of A if and only if f is injective

(d) f(f-'(N)) = N for each subset N of B if and only if f is surjective

Proof (a) Let x E M. To prove x E f -'(f(M)), we need only show

f(x) E f (M). Since x E M, the latter statement is true.

(b) Let y E f (f - '(N)). We claim y E N. Now we know there exists

x E f - '(N) such that y = f (x). Since x E f - '(N), then f (x) E N. Since

y = f (x), we conclude y E N, as desired.

(c) Suppose M = f -I( f(~)) for every subset M of A. To prove f

injective, let x,, x, E A and suppose f (x,) = f (x,). We claim x, = x,.

Letting M = (x,}, we note that f - '( f (M)) c M for this particular set M,

because of our hypothesis. But f (x,) = f (x,) E f (M) so x, E f -I( f (M)).

Since f -'(f(M)) G M = (x,}, we have x, E (x,), so that x, = x,, as

desired. Conversely, suppose f is injective. Let M c A and recall from

(a) that M G f - '( f (M)). To prove equality, we need only show