# Steenrod Algebra Modules#

Author: Robert R. Bruner, Michael J. Catanzaro, Sverre Lunoee–Nielsen, and Koen van Woerden

Let $$p$$ be a prime number. The mod $$p$$ Steenrod algebra $$A$$ is a connected algebra over $$\GF{p}$$, the finite field of $$p$$ elements. All modules presented here will be defined over $$A$$ or one of its sub-Hopf algebras. E.g.:

sage: A = SteenrodAlgebra(p=2)


The constructor of the module class takes as arguments an ordered tuple of degrees and the algebra over which the module is defined, together with an optional set of relations:

sage: from sage.modules.fp_graded.steenrod.module import SteenrodFPModule
sage: F = SteenrodFPModule(A, [0, 1, 7]); F
Free graded left module on 3 generators over mod 2 Steenrod algebra, milnor basis


Denote the module generators of an $$A$$-module $$M$$ by $$g_{d_1}, \ldots, g_{d_N}$$, where subscripts denote their degrees. A homogeneous relation of degree $$n$$ has the form

$\sum_{i=1}^N a_i\cdot g_{d_i} = 0,$

where the homogeneous coefficients $$a_1, \ldots, a_N$$ lie in $$A$$, such that $$\deg(a_i) + \deg(g_{d_i}) = n$$ for $$i = 1, \ldots, N$$. To create a module with relations, the coefficients for each relation is given:

sage: r1 = [Sq(8), Sq(7), 0]   # First relation
sage: r2 = [Sq(7), 0, 1]       # Second relation
sage: M = SteenrodFPModule(A, [0, 1, 7], relations=[r1, r2]); M
Finitely presented left module on 3 generators and 2 relations over mod 2 Steenrod algebra, milnor basis


The resulting module will have three generators in the degrees we gave them:

sage: M.generator_degrees()
(0, 1, 7)


The connectivity of a module over a connected graded algebra is the minimum degree of all its module generators. Thus, if the module is non-trivial, the connectivity is an integer:

sage: M.connectivity()
0


Each module is defined over a Steenrod algebra or some sub-Hopf algebra of it, given by its base ring:

sage: M.base_ring()
mod 2 Steenrod algebra, milnor basis
sage: SteenrodFPModule(SteenrodAlgebra(p=2,profile=(3,2,1)), ).base_ring()
sub-Hopf algebra of mod 2 Steenrod algebra, milnor basis, profile function
[3, 2, 1]


Note

Calling algebra() will not return the desired algebra. Users should use the base_ring() method.

## Module elements#

Module elements are displayed in terms of generators, which by default are called g[degree]:

sage: M.an_element(n=5)
Sq(2,1)*g + Sq(4)*g

sage: e = M.an_element(n=15); e
Sq(0,0,0,1)*g + Sq(1,2,1)*g + Sq(8)*g
sage: e.dense_coefficient_list()
[Sq(0,0,0,1), Sq(1,2,1), Sq(8)]


The generators are themselves elements of the module:

sage: gens = M.generators(); gens
(g, g, g)
sage: gens in M
True


One can produce an element from a given set of algebra coefficients:

sage: coeffs = [Sq(15), Sq(10)*Sq(1,1), Sq(8)]
sage: x = M(coeffs); x
Sq(15)*g + (Sq(4,1,1)+Sq(7,0,1)+Sq(11,1))*g + Sq(8)*g


The module action produces new elements:

sage: Sq(2) * x
Sq(14,1)*g + (Sq(7,1)+Sq(10))*g


Each non-zero homogeneous element has a well-defined degree:

sage: x.degree()
15


but the zero element does not:

sage: zero = M.zero(); zero
0
sage: zero.degree()
Traceback (most recent call last):
...
ValueError: the zero element does not have a well-defined degree


At this point it may be useful to point out that elements are not reduced to a minimal representation when they are created. A normalization can be enforced, however:

sage: g7 = M([0, 0, 1]); g7
g
sage: g7.normalize()
Sq(7)*g
sage: g7 == g7.normalize()
True

sage: m = M([Sq(7), 0, 0])
sage: s = m + g7; s         # m and g7 are related by m = Sq(7)*g = g,
Sq(7)*g + g
sage: s == 0                # so their sum should zero.
True
sage: s.normalize()         # Its normalized form is more revealing.
0


For every integer $$n$$, the set of module elements of degree $$n$$ form a vector space over the ground field $$\GF{p}$$. A basis for this vector space can be computed:

sage: M.basis_elements(7)
(Sq(0,0,1)*g,
Sq(1,2)*g,
Sq(4,1)*g,
Sq(7)*g,
Sq(0,2)*g,
Sq(3,1)*g,
Sq(6)*g)


Note that the third generator $$g_7$$ of degree 7 is apparently missing from the basis above. This is because of the relation $$\operatorname{Sq}^7(g_0) = g_7$$.

A vector space presentation can be produced:

sage: M.vector_presentation(5)
Vector space quotient V/W of dimension 4 over Finite Field of size 2 where
V: Vector space of dimension 4 over Finite Field of size 2
W: Vector space of degree 4 and dimension 0 over Finite Field of size 2
Basis matrix:
[]


Given any element, its coordinates with respect to this basis can be computed:

sage: x = M.an_element(7); x
Sq(0,0,1)*g + Sq(3,1)*g + g
sage: v = x.vector_presentation(); v
(1, 0, 0, 1, 0, 1, 0)


Going the other way, any element can be constructed by specifying its coordinates:

sage: x_ = M.element_from_coordinates((1, 0, 0, 1, 0, 1, 0), 7)
sage: x_
(Sq(0,0,1)+Sq(7))*g + Sq(3,1)*g
sage: x_ == x
True


## Module homomorphisms#

Homomorphisms of $$A$$-modules $$M\to N$$ are linear maps of their underlying $$\GF{p}$$-vector spaces which commute with the $$A$$-module structure. Homomorphisms are required to be homogeneneous but need not be degree zero.

To create a homomorphism, first create the object modeling the set of all such homomorphisms using the function Hom:

sage: Hko = SteenrodFPModule(A, , [[Sq(2)], [Sq(1)]])
sage: homspace = Hom(Hko, Hko); homspace
Set of Morphisms from Finitely presented left module on 1 generator and 2 relations
over mod 2 Steenrod algebra, milnor basis
to Finitely presented left module on 1 generator and 2 relations
over mod 2 Steenrod algebra, milnor basis
in Category of finitely presented graded modules over mod 2 Steenrod algebra, milnor basis


Just as with module elements, homomorphisms are created using the homspace. The only argument is a list of elements in the codomain, giving the images of the module generators of the domain:

sage: gen = Hko.generator(0)  # the generator of the codomain module
sage: values = [Sq(0, 0, 1) * gen]; values
[Sq(0,0,1)*g]
sage: f = homspace(values)


The resulting homomorphism is the one sending the $$i$$-th generator of the domain to the $$i$$-th codomain value given:

sage: f
Module endomorphism of Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
Defn: g |--> Sq(0,0,1)*g


Homomorphisms can be evaluated on elements of the domain module:

sage: v1 = f(Sq(4)*gen); v1
Sq(4,0,1)*g

sage: v2 = f(Sq(2)*Sq(4)*gen); v2
(Sq(3,1,1)+Sq(6,0,1))*g


and they respect the module action:

sage: f(Sq(4)*gen) == Sq(4)*f(gen)
True

sage: f(Sq(2)*Sq(4)*gen) == Sq(2)*Sq(4)*f(gen)
True


Convenience methods exist for creating the trivial morphism:

sage: x = Sq(4)*Sq(7)*gen
sage: x == 0
False
sage: zero_map = homspace.zero(); zero_map
Module endomorphism of Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
Defn: g |--> 0
sage: zero_map(x)
0
sage: zero_map(x).is_zero()
True


as well as the identity endomorphism:

sage: one = Hom(Hko, Hko).identity(); one
Module endomorphism of Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
Defn: g |--> g
sage: one.is_endomorphism()
True
sage: one(x) == x
True
sage: one.is_identity()
True


Any non-trivial homomorphism has a well defined degree:

sage: f.degree()
7


but just as for module elements, the trivial homomorphism does not:

sage: zero_map = homspace.zero()
sage: zero_map.degree()
Traceback (most recent call last):
...
ValueError: the zero morphism does not have a well-defined degree


Any two homomorphisms can be added as long as they are of the same degree:

sage: f1 = homspace([Hko([Sq(0,0,3) + Sq(0,2,0,1)])])
sage: f2 = homspace([Hko([Sq(8,2,1)])])
sage: (f1 + f2).is_zero()
False
sage: f1 + f2
Module endomorphism of Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
Defn: g |--> (Sq(0,0,3)+Sq(0,2,0,1)+Sq(8,2,1))*g


or when at least one of them is zero:

sage: f + zero_map == f
True


but not if they have different degrees:

sage: F = SteenrodFPModule(A, )
sage: b4 = Hom(F, F)([Sq(4) * F.generator(0)])
sage: b8 = Hom(F, F)([Sq(8) * F.generator(0)])
sage: b4 + b8
Traceback (most recent call last):
...
ValueError: morphisms do not have the same degree


sage: (f - f) == 0
True


The restriction of a homomorphism to the vector space of $$n$$-dimensional module elements is a linear transformation:

sage: f_21 = f.vector_presentation(21); f_21
Vector space morphism represented by the matrix:
[1 0 0 0 0 0]
[0 0 0 0 0 0]
[1 0 0 0 0 0]
Domain: Vector space quotient V/W of dimension 3 over Finite Field of size 2 where
V: Vector space of dimension 20 over Finite Field of size 2
W: Vector space of degree 20 and dimension 17 over Finite Field of size 2
Basis matrix:
[1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0]
[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0]
[0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1]
[0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1]
[0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 1]
[0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1]
[0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1]
[0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1]
[0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1]
Codomain: Vector space quotient V/W of dimension 6 over Finite Field of size 2 where
V: Vector space of dimension 35 over Finite Field of size 2
W: Vector space of degree 35 and dimension 29 over Finite Field of size 2
Basis matrix:
29 x 35 dense matrix over Finite Field of size 2


This is compatible with the vector presentations of its domain and codomain modules:

sage: f.domain() is Hko
True
sage: f.codomain() is Hko
True
sage: f_21.domain() is Hko.vector_presentation(21)
True
sage: f_21.codomain() is Hko.vector_presentation(21 + f.degree())
True


Elements in the preimage of a homomorphism can be found:

sage: f.solve(Sq(2)*Sq(4)*Sq(7)*gen)
Sq(0,2)*g

sage: f.solve(Sq(8)*gen) is None
True


Homomorphisms can be composed as expected:

sage: g = homspace([Sq(0, 0, 0, 1)*gen]); g
Module endomorphism of Finitely presented left module on 1 generator and 2 relations
over mod 2 Steenrod algebra, milnor basis
Defn: g |--> Sq(0,0,0,1)*g

sage: g*f
Module endomorphism of Finitely presented left module on 1 generator and 2 relations
over mod 2 Steenrod algebra, milnor basis
Defn: g |--> Sq(0,0,1,1)*g

sage: one = homspace.identity()
sage: f*one == f
True


### Homological algebra#

The category of modules over $$A$$ is Abelian, so kernels, images and cokernels all exist and can be computed using the morphisms.

Note

Methods for morphisms like

compute sub- and quotient modules related to homomorphisms, but they do not return instances of the module class. Rather, they return the natural homomorphisms which connect these modules to the modules that gave rise to them.

For example the function kernel_inclusion() returns an injective homomorphism which is onto the kernel submodule we asked it to compute, while the function cokernel_projection() provides a surjective homomorphism onto the cokernel module.

In each case, getting a reference to the module instance requires calling domain() or codomain() on the returned homomorphism, depending on the case.

Refer to each function’s documentation for specific details.

### Cokernels#

In the following example, we define a cyclic module $$H\ZZ$$ with one relation in two ways: first explicitly, and then as the cokernel of a homomorphism of free modules. We then construct a candidate for an isomorphism and check that it is both injective and surjective:

sage: HZ = SteenrodFPModule(A, , [[Sq(1)]]); HZ
Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis

sage: F = SteenrodFPModule(A, )
sage: j = Hom(F, F)([Sq(1)*F.generator(0)])
sage: coker = j.cokernel_projection() # the natural quotient homomorphism onto the cokernel.
sage: hz = coker.codomain(); hz
Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis

sage: a = Hom(HZ, hz)([hz.generator(0)])
sage: a.is_injective()
True
sage: a.is_surjective()
True


### Kernels#

When computing the kernel of a homomorphism $$f$$, the result is an injective homomorphism into the domain of $$f$$:

sage: k = f.kernel_inclusion(); k
Module morphism:
From: Finitely presented left module on 1 generator and 3 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
Defn: g |--> Sq(0,0,1)*g
sage: k.codomain() == f.domain()
True
sage: k.is_injective()
True

sage: ker = k.domain()
sage: ker
Finitely presented left module on 1 generator and 3 relations over mod 2 Steenrod algebra, milnor basis


We can check that the injective image of $$k$$ is the kernel of $$f$$ by showing that $$f$$ factors as $$h\circ c$$, where $$c$$ is the quotient map to the cokernel of $$k$$, and $$h$$ is injective:

sage: K = k.codomain()        # We want to check that this really is the kernel of f.
sage: coker = k.cokernel_projection()    # coker is the natural map: Hko -> coker(f) with kernel K.
sage: h = Hom(coker.codomain(), Hko)(f.values())
sage: h*coker == f            # Is K contained in ker(f) ?
True
sage: h.is_injective()        # Is ker(f) contained in K ?
True


### Images#

The method image() behaves similarly, returning an injective homomorphism with image equal to the submodule $$\operatorname{im}(f)$$:

sage: i = f.image(); i
Module morphism:
From: Finitely presented left module on 1 generator and 3 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
Defn: g |--> Sq(0,0,1)*g
sage: i.codomain() == f.codomain()
True
sage: i.is_injective()
True


We can check that the injective image of $$i$$ is the image of $$f$$ by lifting $$f$$ over $$i$$, and showing that the lift is surjective:

sage: f_ = f.lift(i); f_
Module morphism:
From: Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 3 relations over mod 2 Steenrod algebra, milnor basis
Defn: g |--> g
sage: i*f_ == f             # Is im(i) contained in im(f) ?
True
sage: f_.is_surjective()    # Is im(f) contained in im(i) ?
True


When a pair of composable homomorphisms $$g\circ f: M\to N\to L$$ satisfy the condition $$g\circ f = 0$$, the sub-quotient $$\ker(g) / \operatorname{im}(f)$$ can be computed and is given by the natural quotient homomorphism with domain $$\ker(g)$$:

sage: f * f == 0        # Does the kernel of f contain the image of f ?
True
sage: K = f.kernel_inclusion()    # k: ker(f) -> Hko
sage: h = f.homology(f) # h: ker(f) -> ker(f) / im(f)
sage: h.codomain()      # This is the homology module.
Finitely presented left module on 1 generator and 4 relations over mod 2 Steenrod algebra, milnor basis


### Free resolutions#

Finally, free resolutions can be computed. These calculations usually take some time to complete, so it is usually a good idea to raise the verbose flag to output progress information.

The following examples are taken from Michael Catanzaro’s thesis where the first version of this software appeared:

sage: res = Hko.resolution(6, verbose=True)
Computing f_1 (1/6)
Computing f_2 (2/6)
Computing using the profile:
(2, 1)
Resolving the kernel in the range of dimensions [1, 8]: 1 2 3 4 5 6 7 8.
Computing f_3 (3/6)
Computing using the profile:
(2, 1)
Resolving the kernel in the range of dimensions [2, 10]: 2 3 4 5 6 7 8 9 10.
Computing f_4 (4/6)
Computing using the profile:
(2, 1)
Resolving the kernel in the range of dimensions [3, 13]: 3 4 5 6 7 8 9 10 11 12 13.
Computing f_5 (5/6)
Computing using the profile:
(2, 1)
Resolving the kernel in the range of dimensions [4, 18]: 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18.
Computing f_6 (6/6)
Computing using the profile:
(2, 1)
Resolving the kernel in the range of dimensions [5, 20]: 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20.


The result of the calculation is a list of all the maps in the resolution:

sage: [f.domain() for f in res]
[Free graded left module on 1 generator over mod 2 Steenrod algebra, milnor basis,
Free graded left module on 2 generators over mod 2 Steenrod algebra, milnor basis,
Free graded left module on 2 generators over mod 2 Steenrod algebra, milnor basis,
Free graded left module on 2 generators over mod 2 Steenrod algebra, milnor basis,
Free graded left module on 3 generators over mod 2 Steenrod algebra, milnor basis,
Free graded left module on 4 generators over mod 2 Steenrod algebra, milnor basis,
Free graded left module on 4 generators over mod 2 Steenrod algebra, milnor basis]

sage: def is_complex(res):
....:     for i in range(len(res)-1):
....:         f = (res[i]*res[i+1])
....:         if not f.is_zero():
....:             return False
....:     return True
....:
sage: is_complex(res)
True

sage: def is_exact(res):
....:     for i in range(len(res)-1):
....:         h = res[i].homology(res[i+1])
....:         if not h.codomain().is_trivial():
....:             return False
....:     return True

sage: is_exact(res)
True

sage: [r.codomain().generator_degrees() for r in res]
[(0,), (0,), (2, 1), (2, 4), (3, 7), (4, 8, 12), (5, 9, 13, 14)]

sage: [r.values() for r in res]
[(g,),
(Sq(2)*g, Sq(1)*g),
(Sq(1)*g, Sq(3)*g + Sq(2)*g),
(Sq(1)*g, Sq(2,1)*g + Sq(3)*g),
(Sq(1)*g, Sq(2,1)*g + Sq(1)*g, Sq(2,1)*g),
(Sq(1)*g,
Sq(2,1)*g + Sq(1)*g,
Sq(2,1)*g + Sq(1)*g,
Sq(2)*g),
(Sq(1)*g,
Sq(2,1)*g + Sq(1)*g,
Sq(2,1)*g + Sq(1)*g,
Sq(0,1)*g + Sq(2)*g)]


## Example: Counting lifts#

In this more elaborate example we show how to find all possible lifts of a particular homomorphism. We will do this in two ways, and as a check of validity, we will compare the results in the end.

We will work with the following modules over the mod 2 Steenrod algebra $$A$$:

\begin{align} H\ZZ &= A/A\cdot \operatorname{Sq}(1)\\ Hko &= A/A\cdot \{\operatorname{Sq}(2), \operatorname{Sq}(1)\} \,. \end{align}

There is a natural projection $$q: H\ZZ\to Hko$$, and a non-trivial endomorphism of degree 28, represented as a degree zero map $$f: \Sigma^{28}Hko\to Hko$$ that we define below.

The problem we will solve is to find all possible homomorphisms $$f': \Sigma^{28}Hko\to H\ZZ$$, making the following diagram into a commuting triangle:

$H\ZZ\xrightarrow{q} Hko \xleftarrow{f} \Sigma^{28} Hko.$

We begin by defining the modules and the homomorphisms $$f$$ and $$q$$. In the following, we let $$L = \Sigma^{28}Hko$$:

sage: from sage.modules.fp_graded.steenrod.module import SteenrodFPModule
sage: A = SteenrodAlgebra(2)
sage: Hko = SteenrodFPModule(A, , [[Sq(2)],[Sq(1)]])
sage: HZ = SteenrodFPModule(A, , [[Sq(1)]])
sage: L = Hko.suspension(28)


The projection:

sage: q = Hom(HZ, Hko)([Hko.generator(0)])
sage: q
Module morphism:
From: Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
Defn: g |--> g


The map to lift over $$q$$:

sage: f = Hom(L, Hko)([Sq(0,2,1,1)*Hko.generator(0)])
sage: f
Module morphism:
From: Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
Defn: g |--> Sq(0,2,1,1)*g

sage: f.is_zero()   # f is non-trivial.
False


We will count the number of different lifts in two ways. First, we will simply compute the vector space of all possible maps $$L \to H\ZZ$$, and then check which of those become $$f$$ when composed with $$q$$:

sage: basis = Hom(L, HZ).basis_elements(0)   # The basis for the vector space of degree 0 maps L -> HZ

sage: from itertools import product
sage: def from_coords(c):
....:     '''
....:     Create a linear combination of the three basis homomorphisms.
....:     '''
....:     return c*basis + c*basis + c*basis

sage: for coords in product([0,1], repeat=3):
....:     print('%s: %s' % (coords, q*from_coords(coords) == f))
(0, 0, 0): False
(0, 0, 1): False
(0, 1, 0): True
(0, 1, 1): True
(1, 0, 0): True
(1, 0, 1): True
(1, 1, 0): False
(1, 1, 1): False


From this we conclude that four out of eight different homomorphisms $$L \to H\ZZ$$ are lifts of $$f$$:

sage: lifts = [from_coords((0,1,0)),
....:          from_coords((0,1,1)),
....:          from_coords((1,0,0)),
....:          from_coords((1,0,1))]
sage: lifts
[Module morphism:
From: Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis
Defn: g |--> Sq(6,5,1)*g,
Module morphism:
From: Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis
Defn: g |--> (Sq(6,5,1)+Sq(18,1,1))*g,
Module morphism:
From: Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis
Defn: g |--> Sq(10,1,0,1)*g,
Module morphism:
From: Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis
Defn: g |--> (Sq(10,1,0,1)+Sq(18,1,1))*g]


Alternatively we can use left-exactness of the functor $$\operatorname{Hom}_A(L, -)$$ to enumerate all possible lifts of $$f$$. Start by finding a single lift of $$f$$ over the projection $$q$$:

sage: fl = f.lift(q); fl
Module morphism:
From: Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis
Defn: g |--> (Sq(4,3,0,1)+Sq(6,0,1,1)+Sq(7,2,0,1)+Sq(10,1,0,1))*g


We verify that fl is indeed a lift:

sage: q * fl == f
True


There is an exact sequence

$0 \to \operatorname{Hom}_A(L, \ker(q)) \xrightarrow{iK_*} \operatorname{Hom}_A(L, H\ZZ) \xrightarrow{q_*} \operatorname{Hom}_A(L, Hko),$

which means that the indeterminacy of choosing a lift for $$f \in \operatorname{Hom}_A(L, Hko)$$ is represented by an element in $$\operatorname{Hom}_A(L,\ker(q))$$. Therefore, we can proceed to count the number of lifts by computing this vector space of homomorphisms:

sage: iK = q.kernel_inclusion()
sage: K = iK.domain()
sage: K.generator_degrees()
(2,)
sage: K.relations()
(Sq(2)*g,)
sage: ind = Hom(L, K).basis_elements(0); len(ind)
2


So now we know that the vector space of indeterminacies is 2-dimensional over the field of two elements. This means that there are four distinct lifts of $$f$$ over $$q$$, and we can construct these by taking the one lift we already found, and add to it all the different elements in the image of $$iK_*$$:

sage: flift = [fl,
....:          fl + iK * ind,
....:          fl + iK * ind,
....:          fl + iK * (ind + ind)]
sage: flift
[Module morphism:
From: Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis
Defn: g |--> (Sq(4,3,0,1)+Sq(6,0,1,1)+Sq(7,2,0,1)+Sq(10,1,0,1))*g,
Module morphism:
From: Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis
Defn: g |--> (Sq(0,7,1)+Sq(3,6,1)+Sq(4,1,3)+Sq(6,0,1,1)+Sq(6,5,1)+Sq(7,0,3))*g,
Module morphism:
From: Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis
Defn: g |--> (Sq(4,3,0,1)+Sq(6,0,1,1)+Sq(7,2,0,1)+Sq(10,1,0,1)+Sq(12,3,1)+Sq(15,2,1)+Sq(18,1,1))*g,
Module morphism:
From: Finitely presented left module on 1 generator and 2 relations over mod 2 Steenrod algebra, milnor basis
To:   Finitely presented left module on 1 generator and 1 relation over mod 2 Steenrod algebra, milnor basis
Defn: g |--> (Sq(0,7,1)+Sq(3,6,1)+Sq(4,1,3)+Sq(6,0,1,1)+Sq(6,5,1)+Sq(7,0,3)+Sq(12,3,1)+Sq(15,2,1)+Sq(18,1,1))*g]


As a test of correctness, we now compare the two sets of lifts. As they stand, it is not obvious that the lists flift and lifts are the same (up to a re-ordering of list elements), so the following comparison is reassuring:

sage: flift == lifts
True
sage: flift == lifts
True
sage: flift == lifts
True
sage: flift == lifts
True