Let h : X −→ Y be defined by
h(x) :=
f(x) if x ∈ F
g
−1
(x) if x ∈ X − F
Now we must prove that h is injective and bijective. Starting
with injectivity, let x1, x2 ∈
X such that h(x1) = h(x2). Assume x1 ∈ F and x2 ∈ X −F. Then h(x1)
= f(x1) ∈ f(F)
and h(x2) = g
−1
(x2) ∈ g
−1
(X − F) = Y − f(F) as follows from question 2 above.
Since
h(x1) = h(x2), and h(x1) = f(x1) ∈ f(F) and h(x2) ∈ Y − f(F), this
is a contradiction
(question 3 why?).
How do I prove this function is not surjective? 3.) Let f: R-R, f(x)-x2+ x+1 and Show that f is not injective and not surjective Justify that g is bijective and find gt. PIR, Show all the wortky) Not Surtechive: fx) RB Surjective: ye(o,oo) hng (g) 8 gon)-es is bijecelive g(x)-ex+s
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Let X = {0, 1, 2} and Y = {0,1,2}. Now we define f={(0,1),(1,0),(2,1)] Please enter your answer as a sum of the following numbers (they are not mutually exclusive): • 1 ifff is a function f : X Y • 2 ifff is a function and it is also injective • 4ifff is a function and it is also surjective This means that your answer can be 0 (not a function), 1 (a function but neither injective or surjective)....
that h(mn ) h ( m)n, h ( ) and that if m < n then h ( m ) < n ( n ) = . Exercise 2.7.4. [Used in Theorem 2.7.1.] Complete the missing part of Step 3 of the proof of Theorem 2.7.1. That is, prove that k is surjective. Exercise 2.7.5. [Used in Theorem 2.7.1.] Let Ri and R2 be ordered fields that satisf We were unable to transcribe this imageWe were unable to transcribe this...
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