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compose -- composition as a pairing on Hom-modules

Synopsis

Description

In the following example we check that the map does implement the composition map $$ \mathrm{Hom}(M,N) \otimes \mathrm{Hom}(N,P) \to \mathrm{Hom}(M,P). $$

i1 : R = QQ[x,y]

o1 = R

o1 : PolynomialRing
 
i2 : M = image vars R ++ R^2

o2 = image | x y 0 0 |
           | 0 0 1 0 |
           | 0 0 0 1 |

                             3
o2 : R-module, submodule of R
 
i3 : f = compose(M,M,M);

o3 : Matrix
 
i4 : H = Hom(M,M);
 
i5 : g = H_{0}

o5 = {0} | 1 |
     {0} | 0 |
     {0} | 0 |
     {0} | 0 |
     {0} | 0 |
     {0} | 0 |
     {0} | 0 |
     {1} | 0 |
     {1} | 0 |
     {1} | 0 |
     {1} | 0 |

                   1
o5 : Matrix H <-- R
 
i6 : h = homomorphism g

o6 = {1} | 1 0 0 0 |
     {1} | 0 1 0 0 |
     {0} | 0 0 0 0 |
     {0} | 0 0 0 0 |

o6 : Matrix M <-- M
 
i7 : f * (g ** g)

o7 = {0} | 1 |
     {0} | 0 |
     {0} | 0 |
     {0} | 0 |
     {0} | 0 |
     {0} | 0 |
     {0} | 0 |
     {1} | 0 |
     {1} | 0 |
     {1} | 0 |
     {1} | 0 |

                   1
o7 : Matrix H <-- R
 
i8 : h' = homomorphism oo

o8 = {1} | 1 0 0 0 |
     {1} | 0 1 0 0 |
     {0} | 0 0 0 0 |
     {0} | 0 0 0 0 |

o8 : Matrix M <-- M
 
i9 : h' === h * h

o9 = true
 
i10 : assert oo

The modules should be defined over the same ring.

See also

Ways to use compose :

For the programmer

The object compose is a method function with options.