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Macaulay2Doc :: inducedMap(Module,Module,Matrix)

inducedMap(Module,Module,Matrix) -- compute the induced map

Synopsis

Description

The modules M and N must both be subquotient modules where M and P have the same ambient module, and N and Q have the same ambient module. If the optional argument Verify is given, check that the result defines a well defined homomorphism.

In this example, the module K2 is mapped via g into K1, and we construct the induced map from K2 to K1.

i1 : R = ZZ/32003[x,y,z]

o1 = R

o1 : PolynomialRing
i2 : g1 = matrix{{x,y,z}}

o2 = | x y z |

             1       3
o2 : Matrix R  <--- R
i3 : g2 = matrix{{x^2,y^2,z^2}}

o3 = | x2 y2 z2 |

             1       3
o3 : Matrix R  <--- R
i4 : K1 = ker g1

o4 = image {1} | -y 0  -z |
           {1} | x  -z 0  |
           {1} | 0  y  x  |

                             3
o4 : R-module, submodule of R
i5 : K2 = ker g2

o5 = image {2} | -y2 0   -z2 |
           {2} | x2  -z2 0   |
           {2} | 0   y2  x2  |

                             3
o5 : R-module, submodule of R
i6 : f = map(ambient K1, ambient K2, {{x,0,0},{0,y,0},{0,0,z}})

o6 = {1} | x 0 0 |
     {1} | 0 y 0 |
     {1} | 0 0 z |

             3       3
o6 : Matrix R  <--- R
i7 : h = inducedMap(K1,K2,f)

o7 = {2} | xy 0  0  |
     {2} | 0  yz 0  |
     {2} | 0  0  xz |

o7 : Matrix
If we omit the first argument, then it is understood to be the target of f, and if we omit the second argument, it is understood to be the source of f.
i8 : h1 = inducedMap(target f,K2,f)

o8 = {1} | -xy2 0    -xz2 |
     {1} | x2y  -yz2 0    |
     {1} | 0    y2z  x2z  |

o8 : Matrix
i9 : h2 = inducedMap(,K2,f)

o9 = {1} | -xy2 0    -xz2 |
     {1} | x2y  -yz2 0    |
     {1} | 0    y2z  x2z  |

o9 : Matrix
i10 : h1 == h2

o10 = true
In this example, we cannot omit the second argument, since in that case the resulting object is not a homomorphism.

See also