Let
be linear systems. Then, the set
is a vector subspace of
and hence an abelian group. Moreover, compositions are bilinear. That is to say, for objects
and
and
the composition maps
are group homomorphisms.
In an additive category, it makes sense to search for kernels and cokernels of morphisms as well as natural (canonical) decompositions of morphisms.
Kernels and Cokernels
The next result is a classical characterization of kernels and cokernels.
Lemma 1. Let be a feedback morphism.
kernels.Morphism defines the kernel of if and only if, for every , the following sequence of abelian groups is exact. kernel diagram.Equivalently, morphism defines the kernel of if there exists a system K and a feedback morphism ι making the top triangle commutative, and for each linear system Γ and each morphism j making the bottom triangle commutative, there exists a unique feedback morphism making the left triangle commutative.
cokernels.Morphism defines the cokernel of if and only if, for every , the following sequence of abelian groups is exact. cokernel diagram.Equivalently, morphism defines the cokernel of if there exists a system C and a feedback morphism 𝔠 making the top triangle commutative, and for each linear system Γ and each morphism j making the bottom triangle commutative, there exists a unique feedback morphism making the right triangle commutative
In the following, the kernel (respectively, cokernel) of a feedback map in is denoted by (respectively, ) while (respectively, ) are often used, by abuse of notation, for the kernel (respectively, cokernel) of linear map in -Vect, where is the functor defined in the above Note 1.
Lemma 2. Let be a feedback morphism. Assume that its kernel does exist in . Then, injects into (we often write ).
On the other hand, maps onto (we often write maps onto ).
Proof. It is immediate because is the kernel of in -Vect and . The dual argument works for cokernels. □
Corollary 1. Let be a feedback morphism.
(i) If is an injective map, then .
(ii) If is a surjective map, then .
Theorem 1. Category has all cokernels.
Proof. Let
be a feedback morphism. Assume
and
. Set bases in
V and
so that the matrix of
(as a linear map between vector spaces) is of the Hermite form. Obtain the block matrices associated with
f and
, as well as the basis of subspaces
B and
. Then, the above systems and morphism are transformed into:
where
.
It is important to note here that, since is a feedback morphism, it follows that . Hence, one has that the image of block is contained in the image of block .
Consider the cokernel diagram:
A cokernel makes the upper triangle commutative and for all systems and morphism making the lower triangle commutative, there exists a unique feedback morphism making the right triangle commutative.
Define the linear system and linear map .
- (i)
- (ii)
Moreover, is a feedback morphism. On the other hand, it is clear that
Now, set any linear system and note that:
- (iv)
A morphism must be on the block form in order to assure
- (v)
and because is a feedback morphism
We claim that is the unique feedback morphism making the right triangle commutative.
- (vi)
(by iv) yields
- (vii)
by (v).
Hence, is a feedback morphism. It is quite clear that is the unique solution to equation , which is, in matrix form, □
Next, some sufficient conditions for the existence of kernels in the category of linear systems are given.
Theorem 2 (Sufficient condition of kernels).
Let be a feedback morphism. If the kernel of as linear map is f-invariant (i.e., ), then the natural inclusion defines a feedback morphism and Proof. Consider the kernel diagram
Then,
is a well defined feedback morphism and it is unique, making the left triangle commutative □
Next, we give an example of the above condition of existence of kernels in :
Example 4. Consider the linear systems over and given by The linear map verifies the following properties:
is a feedback morphism because:
- (i)
- (ii)
Thus, is a feedback morphism. and . Then, the kernel is Note 2. The above sufficient condition to the existence of kernels in is, in general, not necessary. The feedback map given byverifies: - (i)
(trivial).
- (ii)
is not f-invariant because
- (iii)
However, the morphism has a kernel in . In fact, in because of any feedback morphism j in the below diagramneeds to be .
Note also that , while . That is to say, is in general not isomorphic to .