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6 Pages

### m200lxiv

Course: M 200, Fall 2009
School: Maryland
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Word Count: 1144

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Assignment: Math200 7.5: 5,6,7,8,15-21(odd),24,25 7.6: 1-29 odd Lecture XI Feb 28 2002 Note that our proof (last page, lecture XIII) is also a proof of thm 7.5.2 Composition is Associative (exercise 19) Proof: Let f: Z W g: Y Z h: X Y , WLOG (Without Loss of Generality) Assume h, g, f are onto (i.e., h(X) =Y, g(Y) =Z f(Z) =W ) and that Domf = Z, Domg = Y, Domh = X X h Y g Z f W f g h is well defined iff...

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Assignment: Math200 7.5: 5,6,7,8,15-21(odd),24,25 7.6: 1-29 odd Lecture XI Feb 28 2002 Note that our proof (last page, lecture XIII) is also a proof of thm 7.5.2 Composition is Associative (exercise 19) Proof: Let f: Z W g: Y Z h: X Y , WLOG (Without Loss of Generality) Assume h, g, f are onto (i.e., h(X) =Y, g(Y) =Z f(Z) =W ) and that Domf = Z, Domg = Y, Domh = X X h Y g Z f W f g h is well defined iff composition is associative! Then based on the above assumption we can say with confidence: z Z y Y : g(y) = z and y Y x X : h(x) = y and w W z Z : f(z) = w Hence to prove associativity we must show that (f g) h = f (g h) = f g h which is an equation between functions and is therefore equivalent to showing that: x X w W :((f g) h)(x) = (f (g h))(x) = w (Interpreted as saying there exists an element w such that the LHS and RHS of above expression map to this same point,' and this must be true for all x! ) We use the definition of composition to establish LHS = RHS of above equality: x X :((f g) h)(x) x X LHS: RHS: Because h is onto Because g is onto :(f (g h))(x) :f (g(h(x))) f(g(h(x))) = f(g(y)) = f(z) = w (f g)(h(x)) = (f g)(y) f(g(y)) = f(z) = w Hence iterated composition is associative and therefore any function can be considered the composition of a string of arbitrarily many functions: i.e., f = f 1 f 2 f 3... f n is well-defined on Xn+1f1( Xn ... X3 fn-1( X2 fn(X1))...) This cumbersome expression is the (pedantic) result of ensuring the domains of each and every internal function=range of previous function. There are more elegant ways to express this string (for one thing, note: since fn(X1)) X2, then obviously: X2 fn(X1) = fn(X1). Nevertheless the cumbersome expression is valuable nsofar as it reminds us the domain of next function must be intersected with range of previous function!) ** Of course, all this can be avoided if we assume (like in our proof) that each and every one of these functions are onto** I.) Composition Respects bijections (the composition of two invertible maps is invertible) f: Y Z g: X Let Y be 1-1 correspondences: Then because f,g are 1-1: f(y1) = f(y2) y1 = y2 and g(x1) = g(x2) x1 = x2 Because g is onto1, we know there exist x1 , x2 such that y1 = g(x1) and y2 = g(x2) Therefore: F f(g(x1)) = f(g(x2)) g(x1)= g(x2) x1 = x2 Or by syllogism, f g is 1-1 Because f,g are onto: z Z y Y : f(y) = z and y Y x X : g(x) = y x X : f(g(x)) = z Therefore (by syllogism): z Z Or f g is onto Is: H: R \{1} R \{1} equal to its inverse? H(x) = x + 1 1 The textbook is less pedantic: Epp shows the composition of two 1-1 maps is 1-1 by (implicitly) restricting herself to the case where g(x)=y, i.e. where: Domf=Y =g(X). Of course this assumption is always valid, since the statement is a conditional: the premise is f(g(x1)) = f(g(y2)) which of course obviously holds in the restriction of Domf=Y =g(X). In the above proof, I merely make this assumption explicit to argue that f(y1) = f(y2) f(g(x1))= f(g(x2)) g(x2) g(x1)= x1 = x2 holds true without exception. This without exception' issue is only important when considering 11correspondences (they must be everywhere 1-1 on Domg = X) x - 1 We need to show: H(H-1(y)) = H(H(y)) = y OR H-1(H(x)) = H(H(x)) = x Exercise 16: If gf is onto, need both f and g be onto? Because gf is onto: z Z x X : f(g(x)) = z Re-name: y = g(x) Then: z Z y Y : f(y) = z Therefore f must be onto. ...But g need not be onto Counterexample: Problem 24: For all A X, f -1(f(A)) = A By definition: f -1(f(A)) = { x X: f(x) f(A) } Observe that: x A x f -1(f(A)) , i.e. A f -1(f(A)) But the converse is false: Counterexample: Let f: R where f(x)=x2 R Let A= [0,1]. Then: f(A) = A = [0,1]. But f -1(f(A)) = [-1,1] In fact: f -1(f(A)) = A iff f is invertible (i.e. a 1-1 correspondence) Equivalence Two sets are equivalent: X them. Y, whenever: X = Y There exists a 1-1 correspondence between Proof that Z+ Z . Let's construct a function between these two sets that's not the same function specified in page 415 (text) (after all there are a non-unique number of 1-1correspondences between two equivalnet sets) Consider: f = n-2 n=2m (for m >0) 2 - n/2 n=2k+1 (for k > 0) (Note: we could also conceivably consider a function g: Z + . The relation is symmetric precisely Z because the function should be invertible. Observe that: for k=1,2,3,... the top case maps to 0,1,2,3,... and the bottom case maps to -1,-2,-3,... Therefore, for all n > 0, case 1 > 0 and case2 < 0. f is 1-1: Let f(n1) = f(n2) . Then the only allowed cases are that the two cases equal one another or the two botfom cases equal one another (why?) The top case is trivial to show 1-1. The bottom case amounts to: . n1 /2 =. n2 /2 . (2k1 + 1)/2 = . (2k2 + 1)/2 . k1 + 1/2...

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