Commutative Algebra/Kernels, cokernels, products, coproducts

Kernels
Note that here, we don't see kernels only as subsets, but rather as an object together with a morphism. This is because in the category of groups, for example, we can take the morphism just by inclusion. Let me explain.

Example 3.2:

In the category of groups, every morphism has a kernel.

Proof:

Let $$G, H$$ be groups and $$\varphi: G \to H$$ a morphism (that is, a group homomorphism). We set
 * $$o_k := \{g \in G: \varphi(g) = 0\}$$

and
 * $$k : o_k \to G, k(g) = g$$,

the inclusion. This is indeed a kernel in the category of groups. For, if $$\theta: K \to G$$ is a group homomorphism such that $$\varphi \circ \theta = 0$$, then $$\theta$$ maps wholly to $$o_k$$, and we may simply write $$\theta = k \circ \theta$$. This is also clearly a unique factorisation.

For kernels the following theorem holds:

Proof:

Let $$k \circ s = k \circ t$$. The situation is depicted in the following picture:


 * Kernel-thm1.svg

Here, the three lower arrows depict the general property of the kernel. Now the morphisms $$k \circ s$$ and $$k \circ t$$ are both factorisations of the morphism $$k \circ s$$ over $$k$$. By uniqueness in factorisations, $$s = t$$.

Kernels are essentially unique:

Proof:

From the first property of kernels, we obtain $$f \circ k = 0$$ and $$f \circ \tilde k = 0$$. Hence, the second property of kernels imply the commutative diagrams
 * Kernel-thm2-1.svg and Kernel-thm2-2.svg.

We claim that $$k'$$ and $$\tilde k'$$ are inverse to each other.
 * $$\tilde k k' \tilde k' = k \tilde k' = \tilde k = \tilde k 1_{o_{\tilde k}}$$ and $$k \tilde k ' k' = \tilde k k' = k = k 1_{o_k}$$.

Since both $$k$$ and $$\tilde k$$ are monic by theorem 3.3, we may cancel them to obtain
 * $$k' \tilde k' = 1_{o_{\tilde k}}$$ and $$\tilde k ' k' = 1_{o_k}$$,

that is, we have inverse arrows and thus, by definition, isomorphisms.

Cokernels
An analogous notion is that of a cokernel. This notion is actually common in mathematics, but not so much at the undergraduate level.

Again, this notion is just a generalisation of facts observed in "everyday" categories. Our first example of cokernels shall be the existence of cokernels in Abelian groups. Now actually, cokernels exist even in the category of groups, but the construction is a bit tricky since in general, the image need not be a normal subgroup, which is why we may not be able to form the factor group by the image. In Abelian groups though, all subgroups are normal, and hence this is possible.

Example 3.6:

In the category of Abelian groups, every morphism has a cokernel.

Proof:

Let $$G, H$$ be any two Abelian groups, and let $$\varphi: G \to H$$ be a group homomorphism. We set
 * $$o_u := H / \operatorname{im} \varphi$$;

we may form this quotient group because within an Abelian group, all subgroups are normal. Further, we set
 * $$u: H \to H / \operatorname{im} \varphi, u(h) = h + \operatorname{im} \varphi$$,

the projection (we adhere to the custom of writing Abelian groups in an additive fashion). Let now $$\eta: H \to I$$ be a group homomorphism such that $$\eta \circ \varphi = 0$$, where $$I$$ is another Abelian group. Then the function
 * $$\eta': H / \operatorname{im} \varphi \to I, \eta'(h + \operatorname{im} \varphi) := \eta(h)$$

is well-defined (because of the rules for group morphisms) and the desired unique factorisation of $$h$$ is given by $$h = \eta' \circ u$$.

Proof:

Let $$f$$ be a morphism and $$u$$ a corresponding cokernel. Assume that $$t \circ u = s \circ u$$. The situation is depicted in the following picture:
 * Cokernel-thm1.svg

Now again, $$t \circ u \circ f = 0$$, and $$t \circ u$$ and $$s \circ u$$ are by their equality both factorisations of $$t \circ u$$. Hence, by the uniqueness of such factorisations required in the definition of cokernels, $$s = t$$.

Proof:

Once again, we have $$u \circ f = 0$$ and $$\tilde u \circ f = 0$$, and hence we obtain commutative diagrams
 * Cokernel-thm2-1.svg and Cokernel-thm2-2.svg.

We once again claim that $$u'$$ and $$\tilde u'$$ are inverse to each other. Indeed, we obtain the equations
 * $$u' \tilde u' u = u' \tilde u = u = 1_{o_u} u$$ and $$\tilde u' u' \tilde u = \tilde u' u = \tilde u = 1_{o_{u'}} \tilde u$$

and by cancellation (both $$u$$ and $$\tilde u$$ are epis due to theorem 8.7) we obtain
 * $$u' \tilde u' = 1_{o_u}$$ and $$\tilde u' u' = 1_{o_{u'}}$$

and hence the theorem.

Interplay between kernels and cokernels
Proof:

$$k = \ker f$$ means
 * ker-simple.svg.

We set $$q := \operatorname{coker} k$$, that is,
 * coker-simple2.svg.

In particular, since $$f k = 0$$, there exists a unique $$f'$$ such that $$f = f' q$$. We now want that $$k$$ is a kernel of $$p$$, that is,
 * ker-simple2.svg.

Hence assume $$ql = 0$$. Then $$f l = f' q l = 0$$. Hence, by the topmost diagram (in this proof), $$l = k l'$$ for a unique $$l'$$, which is exactly what we want. Further, $$qk = 0$$ follows from the second diagram of this proof.

Proof:

The statement that $$q$$ is the cokernel of $$r$$ reads
 * Coker-simple.svg.

We set $$k := \ker q$$, that is
 * ker-simple2.svg.

In particular, since $$qr = 0$$, $$r = kr'$$ for a suitable unique morphism $$r'$$. We now want $$q$$ to be a cokernel of $$k$$, that is,
 * coker-simple2.svg.

Let thus $$mk = 0$$. Then also $$mr = mkr' = 0$$ and hence $$m$$ has a unique factorisation $$m = m'q$$ by the topmost diagram.

The equation
 * $$\ker f = \ker(\operatorname{coker}(\ker f))$$

is to be read "the kernel of $$f$$ is a kernel of any cokernel of itself", and the same for the other equation with kernels replaced by cokernels and vice versa.

Proof:

$$k := \ker f$$ is a morphism which is some kernel. Hence, by theorem 3.9
 * $$k = \ker(\operatorname{coker}(k))$$

(where the equation is to be read "$$k$$ is a kernel of any cokernel of $$k$$"). Similarly, from theorem 3.10
 * $$q = \operatorname{coker}(\ker(q))$$,

where $$q := \operatorname{coker} f$$.

Products
Example 3.13:

Coproducts
Example 3.17:

Biproducts
Example 3.21:

Within the category of Abelian groups, a biproduct is given by the product group; if $$G, H$$ are Abelian groups, set the product group of $$G$$ and $$H$$ to be
 * $$G \times H$$,

the cartesian product, with component-wise group operation.

Proof: