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A isomorphism (Isomorphismus) is a bijective homomorphism.
h: A ->hom(f,f') B : <=> h: A -> B /\ (exists n in N: f: An -> A /\ f': Bn -> B /\ (forall x in An: h(f(x0, ..., xn-1)) = f'(h(x0), ..., h(xn-1)))).
h: A ->iso(f,f') B : <=> h: A ->hom(f,f') B /\ h: A ->bijective B.
A homomorphism makes the following diagram commute:
In other words, it does not matter whether we first perform the operation f in A and let h map the result to B or whether we let h map the arguments of f to B and apply f' to them. In this sense, h embeds A into B and their operations are structurally equal. If h is an isomorphism, then this embedding works in both directions: A and B become "identical twins" that cannot be distinguished by the behavior of f and f'.
h(x) := <x, 0>.Then we have, for all x in N and y in N,
i.e., h is a homomorphism from N to Z (for the operations + and *).
h(x+Ny) = h(x) +Z h(y); h(x*Ny) = h(x) *Z h(y)
h(x) := x/1Z.Then we have, for all x in Z and y in Z,
i.e., h is a homomorphism from Z to Q (for operations +, -, and *).
h(x+Zy) = h(x) +Q h(y); h(x-Zy) = h(x) -Q h(y); h(x*Zy) = h(x) *Q h(y)
We have for all x in List(T) and y in List(T)
length(append(x, y)) = length(x) + length(y)i.e., "length" is a homomorphism from "List(T)" to N with respect to "append" and +.
i.e., polynomial evaluation is a homomorphism from the set of polynomials to R (for operations +, -, *).
(x + y)[a] = x(a) +R y(a) (x - y)[a] = x(a) -R y(a) (-x)[a] = -Rx(a) (x * y)[a] = x(a) *R y(a)
As the previous example suggests, we have a homomorphism from each number domain introduced in Chapter Numbers to its successor. Actually, each domain is isomorphic to some subset of its successor, i.e.,
N subset ' Z subset ' Q subset ' R subset ' Cwhere
A subset ' B : <=> exists h, B' subset B: h: A ->iso(OA,O'B) B'.Here OA denotes the considered operations on A, O'B denotes the corresponding operations on B' and h: A ->iso(OA,O'B) B' states that h is an isomorphism between A and B for each operation pair.
Logic Evaluator We demonstrate that (a finite subset of) N is isomorphic to a subset of Z:
Isomorphisms are also useful to demonstrate the equivalence of different constructions as shown below.
Every element x+yi of C can be visualized as a point in the plane whose Cartesian coordinates (kartesische Koordinaten) are denoted by the real part and by the imaginary part of the complex number, respectively:
In polar coordinates (Polarkoordinaten), each point is represented by a pair <r, alpha > where r denotes the distance of the point from the origin and alpha denotes the angle to the horizontal axis:
Thus we can define a "polar" variant of C
C' := R x [0,2pi[assuming that angles are uniquely expressed in "radians" (i.e., pi = 180 o ).
A translation from the new domain into the original one is given by
where "cos" and "sin" represent cosine and sine, respectively.
cartesian: C' -> C cartesian(z) = z0*cos(z1)+z0*sin(z1)i.
The other direction is a bit more complicated:
where "arctan" is the inverse ("arcus") tangent. Since the inverse tangent does not determine the result uniquely, we have to make a case distinction and normalize the result by the following auxiliary function:
polar: C -> C' polar(z) = (sqrt(z02+z12), alpha) where alpha= if z0 = 0 then if z1 > 0 then pi /2 else 3 pi /2 else let a = arctan(z1/z0): shift(if z0 >= 0 then a else pi -a)
shift: R -> [0,2 pi [ shift(a) := (such b: b in [0,2 pi [ /\ exists i in Z: a-b = 2 pi i)
One can now show
i.e., "cartesian" and "polar" are bijections.
cartesian o polar = 1C, polar o cartesian = 1C'
We may define the following "multiplication" on C':
x *C' y := <x0*y0, shift(x1+y1)>We then have
i.e., C and C' are isomorphic with respect to their respective notions of multiplication.
cartesian: C ->iso(*C, *C') C' polar: C' ->iso(*C', *C) C
forall x in C', y in C': cartesian(x *C' y) = cartesian(x) *C cartesian(y).Take arbitrary x in C' and y in C'. We then have
(*) holds because of the definition of "shift" and because, for every x in R,
cartesian(x *C' y) = cartesian(x0y0, shift(x1+y1)) = x0y0cos(shift(x1+y1)) + (x0y0sin(shift(x1+y1)))i = (*) x0y0cos(x1+y1) + (x0y0sin(x1+y1))i.
which we assume as granted knowledge.
sin(x+2 pi ) = sin(x), cos(x+2 pi ) = cos(x)
We also have
We assume the knowledge
cartesian(x) *C cartesian(y) = (x0cos(x1) + x0sin(x1)i) *C (y0cos(y1) + y0sin(y1)i) = (x0y0cos(x1)cos(y1) - x0y0sin(x1)sin(y1)) + (x0y0cos(x1)sin(y1) + x0y0sin(x1)cos(y1))i = x0y0(cos(x1)cos(y1) - sin(x1)sin(y1)) + x0y0(cos(x1)sin(y1) + x0y0sin(x1)cos(y1))i
as granted and are therefore done.
cos(x1+y1) = cos(x1)cos(y1) - sin(x1)sin(y1) sin(x1+y1) = cos(x1)sin(y1) + sin(x1)cos(y1)
We may also define corresponding notions of addition, subtraction, and division in C' such that they are isomorphic to their counterparts in C.
Consequently, we may operate in C or in C', whatever is more convenient, and translate the results into the other direction. In particular, if the computation of some operation is rather awkward in one domain, it may be simpler to translate the arguments into the other domain, perform the operation there, and translate the result back into the original domain. This is in particular true for the computation of complex roots.
sqrt( ): (N x C') -> C' sqrtn(z) := <sqrtn(z0), z1/n>.
forall n in N>0, z in C': let r = sqrtn(z): (forall s in C': z = sn <=> exists i in N: s = <r0, shift(r1+2 pi i/n)>).
In other words, the n-th roots of a complex number are on the same circle and have equal distance from each other as shown in the following diagram for the cube root of <r, alpha >.
Since roots can be computed in C' most easily, one usually does not bother to introduce this concept in C but simply defines:
sqrt(`): (N>0 x C) -> C sqrtn(z) := cartesian(sqrtn(polar(z))).
Vice versa, addition and subtraction are done more easily in C than in C':
+C': (C' x C') -> C' x +C' y := polar(cartesian(x) +C cartesian(y)); -C': (C' x C') -> C' x -C' y := polar(cartesian(x) +C cartesian(y)).