Morphisms of free modules.

AUTHOR:

• William Stein

TESTS:

sage: V = ZZ^2; f = V.hom([V.1,-2*V.0])
sage: loads(dumps(f))
Free module morphism defined by the matrix
[ 0  1]
[-2  0]
Domain: Ambient free module of rank 2 over the principal ideal domain ...
Codomain: Ambient free module of rank 2 over the principal ideal domain ...
sage: loads(dumps(f)) == f
True

class sage.modules.free_module_morphism.FreeModuleMorphism(parent, A)

Bases: sage.modules.matrix_morphism.MatrixMorphism

change_ring(R)

Change the ring over which this morphism is defined. This changes the ring of the domain, codomain, and underlying matrix.

EXAMPLES:

sage: V0 = span([[0,0,1],[0,2,0]],ZZ); V1 = span([[1/2,0],[0,2]],ZZ); W = span([[1,0],[0,6]],ZZ)
sage: h = V0.hom([-3*V1.0-3*V1.1, -3*V1.0-3*V1.1])
sage: h.base_ring()
Integer Ring
sage: h
Free module morphism defined by the matrix
[-3 -3]
[-3 -3]...
sage: h.change_ring(QQ).base_ring()
Rational Field
sage: f = h.change_ring(QQ); f
Free module morphism defined by the matrix
[-3 -3]
[-3 -3]
Domain: Vector space of degree 3 and dimension 2 over Rational Field
Basis ...
Codomain: Vector space of degree 2 and dimension 2 over Rational Field
Basis ...
sage: f = h.change_ring(GF(7)); f
Free module morphism defined by the matrix
[4 4]
[4 4]
Domain: Vector space of degree 3 and dimension 2 over Finite Field of ...
Codomain: Vector space of degree 2 and dimension 2 over Finite Field of ...

inverse_image(V)

Given a submodule V of the codomain of self, return the inverse image of V under self, i.e., the biggest submodule of the domain of self that maps into V.

EXAMPLES:

We test computing inverse images over a field:

sage: V = QQ^3; W = span([[1,2,3],[-1,2,5/3]], QQ)
sage: phi = V.hom(matrix(QQ,3,[1..9]))
sage: phi.rank()
2
sage: I = phi.inverse_image(W); I
Vector space of degree 3 and dimension 2 over Rational Field
Basis matrix:
[   1    0    0]
[   0    1 -1/2]
sage: phi(I.0) in W
True
sage: phi(I.1) in W
True
sage: W = phi.image()
sage: phi.inverse_image(W) == V
True

We test computing inverse images between two spaces embedded in different ambient spaces.:

sage: V0 = span([[0,0,1],[0,2,0]],ZZ); V1 = span([[1/2,0],[0,2]],ZZ); W = span([[1,0],[0,6]],ZZ)
sage: h = V0.hom([-3*V1.0-3*V1.1, -3*V1.0-3*V1.1])
sage: h.inverse_image(W)
Free module of degree 3 and rank 2 over Integer Ring
Echelon basis matrix:
[0 2 1]
[0 0 2]
sage: h(h.inverse_image(W)).is_submodule(W)
True
sage: h(h.inverse_image(W)).index_in(W)
+Infinity
sage: h(h.inverse_image(W))
Free module of degree 2 and rank 1 over Integer Ring
Echelon basis matrix:
[ 3 12]

We test computing inverse images over the integers:

sage: V = QQ^3; W = V.span_of_basis([[2,2,3],[-1,2,5/3]], ZZ)
sage: phi = W.hom([W.0, W.0-W.1])
sage: Z = W.span([2*W.1]); Z
Free module of degree 3 and rank 1 over Integer Ring
Echelon basis matrix:
[    2    -4 -10/3]
sage: Y = phi.inverse_image(Z); Y
Free module of degree 3 and rank 1 over Integer Ring
Echelon basis matrix:
[  6   0 8/3]
sage: phi(Y) == Z
True

lift(x)

Given an element of the image, return an element of the codomain that maps onto it.

EXAMPLE:

sage: X = QQ**2
sage: V = X.span([[2, 0], [0, 8]], ZZ)
sage: W = (QQ**1).span([[1/12]], ZZ)
sage: f = V.hom([W([1/3]), W([1/2])], W)
sage: f.lift([1/3])
(8, -16)
sage: f.lift([1/2])
(12, -24)
sage: f.lift([1/6])
(4, -8)
sage: f.lift([1/12])
...
ValueError: element is not in the image
sage: f.lift([1/24])
...
TypeError: element (= [1/24]) is not in free module

This works for vector spaces, too:

sage: V = VectorSpace(GF(3), 2)
sage: W = VectorSpace(GF(3), 3)
sage: f = V.hom([W.1, W.1 - W.0])
sage: f.lift(W.1)
(1, 0)
sage: f.lift(W.2)
...
ValueError: element is not in the image
sage: w = W((17, -2, 0))
sage: f(f.lift(w)) == w
True

sage.modules.free_module_morphism.is_FreeModuleMorphism(x)

EXAMPLES:

sage: V = ZZ^2; f = V.hom([V.1,-2*V.0])
sage: sage.modules.free_module_morphism.is_FreeModuleMorphism(f)
True
sage: sage.modules.free_module_morphism.is_FreeModuleMorphism(0)
False