PlanetMath: abelian groups form an abelian category, proof that

Proof. We will justify all the axioms.

(Axiom 1.) Suppose and are abelian groups. We need to show that ${\mathrm{Hom}}(A,B)$ has the structure of an abelian group. Suppose $f\colon A\to B$ and $g\colon A\to B$ are elements of ${\mathrm{Hom}}(A,B)$ , and define their sum $f+g\colon A\to B$ by the rule

$\displaystyle (f+g)(x) = f(x) + g(x)$

for any $x\in A$ . This operation inherits the commutativity and associativity of the addition in

. Moreover, the function

is a homomorphism. To see this, suppose

and

are in

. Then

$\displaystyle (f+g)(x+y)$	$\displaystyle = f(x+y) + g(x+y) = f(x) + g(x) + f(y) + g(y)$
	$\displaystyle = (f+g)(x) + (f+g)(y).$

The identity in ${\mathrm{Hom}}(A,B)$ is the constant zero function $0\colon A\to B$ , since for any $x\in A$ ,

$\displaystyle (f+0)(x) = f(x) + 0(x) = f(x) + 0 = f(x).$

Thus ${\mathrm{Hom}}(A,B)$ is an abelian group.

Now we show that composition of morphisms distributes over addition in ${\mathrm{Hom}}(\cdot,\cdot)$ . Suppose we are given a diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ A \ar[r]^f & B\ar@/^/[r]^g\ar@/_/[r]_h & C\ar[r]^k & D } } \end{xy}$

of abelian groups. We claim that

and

. Since the proofs are similar, we prove only the first identity. Let $x\in A$ . Then

$\displaystyle ((g+h)f)(x)$	$\displaystyle = (g+h)(f(x)) = g(f(x)) + h(f(x))$
	$\displaystyle = (gf)(x) + (hf)(x) = (gf+hf)(x).$

Thus $\mathbf{Ab}$ satisfies Axiom 1.

(Axiom 2.) The trivial group 0 is a zero object in $\mathbf{Ab}$ . It is initial because there exists a unique morphism $0\to A$ for any abelian group , and it is terminal because there exists a unique morphism $A\to 0$ for any abelian group . In both cases the morphism is the constant zero function.

(Axiom 3.) The Cartesian product of two abelian groups is a categorical direct product. Since the Cartesian product of two abelian groups is again abelian, it follows that $\mathbf{Ab}$ has finite products.

(Axiom 4.) Now we show that $\mathbf{Ab}$ has kernels and cokernels. Let $f\colon A\to B$ be a morphism. Define a subset $K\subset A$ by $K = \{ x\in A\colon f(x) = 0\}$ , and let $i\colon K\to A$ be inclusion. The set is the group-theoretic kernel of , hence a normal subgroup of . Since is abelian, so is . The inclusion $i\colon K\to A$ is a morphism in $\mathbf{Ab}$ . Moreover, by construction we have that .

So suppose that $j\colon L\to A$ is a morphism in $\mathbf{Ab}$ such that . We need to show that the diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ & L\ar[d]^j\ar@{.>}[dl]_{\tilde{\j}} & \ K\ar[r]_i & A\ar[r]_f & B } } \end{xy}$

has a unique filler $\tilde{\j}$ . Let $x\in L$ . Since

, it follows that $j(x)\in K$ . So define $\tilde{\j}(x)=j(x)$ . Since the inclusion $i\colon K\to A$ is injective, any alternative choice for $\tilde{\j}$ fails to make the diagram commute. So $\mathbf{Ab}$ has kernels.

Now we construct a cokernel for . Define a subset $I\subset B$ by

$\displaystyle I = \{ f(x) \colon x\in A\}.$

Then

is a subgroup of the abelian subgroup

, so we may form the quotient group

. Thus

is the group-theoretic cokernel of

. Define a group homomorphism $p\colon B\to C$ by

. Since

is a quotient of an abelian group, it is abelian, so $p\colon B\to C$ is a morphism in $\mathbf{Ab}$ . Moreover, by construction we have that

So suppose that $q\colon A\to D$ is a morphism in $\mathbf{Ab}$ such that . We need to show that the diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ A\ar[r]^f & B\ar[r]^p\ar[d]_q & C\ar@{.>}[dl]^{\tilde{q}} \ & D } } \end{xy}$

has a unique filler. Let $x+I\in C$ . Suppose

is another representative in

for

, that is, that

. Then $x-y\in I$ , that is, there is some $w\in A$ such that

. By assumption,

, so

. So there is a well-defined function $\tilde{q}\colon C\to D$ defined by $\tilde{q}(x+I) = q(x)$ . Moreover, for any

and

$\displaystyle \tilde{q}(x+y+I) = q(x+y) = q(x) + q(y) = \tilde{q}(x+I) +\tilde{q}(y+I),$

so $\tilde{q}$ is a morphism in $\mathbf{Ab}$ . But any filler for the diagram must be defined in exactly the way we have defined it in order for the diagram to commute. Hence $\mathbf{Ab}$ has cokernels.

(Axiom 5.) Suppose $f\colon A\to B$ is a monomorphism, and let $p\colon B\to{\mathrm{coker}}f$ be a cokernel of . We must show that is a kernel of . Since is a cokernel of , we know that . Now we must show that if $j\colon L\to B$ is any morphism in $\mathbf{Ab}$ such that , then the diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ & L\ar[d]^j\ar@{.>}[dl]_{\tilde{\j}} & \ A\ar[r]_f & B\ar[r]_<<<p & {\mathrm{coker}}f } } \end{xy}$

has a unique filler $\tilde{\j}$ . So suppose $x\in L$ . Since

, it follows by the construction of the cokernel given above that there is some $y\in A$ such that

. Since

is monomorphism,

is injective, so this

is unique. We may therefore define $\tilde{\j}$ by the formula $\tilde{\j}(x) = f^{-1}(j(x))$ . Hence

is a kernel for the cokernel of

(Axiom 6.) Suppose $f\colon A\to B$ is an epimorphism, and let $i\colon \ker f\to A$ be a kernel of . We must show that is a cokernel of . Since is a kernel of , we know that . Now we must show that if $q\colon A\to D$ is any morphism in $\mathbf{Ab}$ such that , then the diagram

$\displaystyle \begin{xy} *!C\xybox{ \xymatrix{ \ker f\ar[r]^i & A\ar[r]^f\ar[d]_q & B\ar@{.>}[dl]^{\tilde{q}} \ & D & } } \end{xy}$

has a unique filler $\tilde{q}$ .

To prove the existence of $\tilde{q}$ , first recall that epimorphisms in $\mathbf{Ab}$ are surjections. Let $z\in B$ . Suppose . Then is in $\ker f$ . By assumption, , so this implies that . So we may define a morphism $\tilde{q}\colon B\to D$ in $\mathbf{Ab}$ by the formula $\tilde{q}(z) = q(x)$ , where is any element of $f^{-1}(z)$ . Moreover, $\tilde{q}f = q$ by construction.

To prove the uniqueness of $\tilde{q}$ , suppose that $\hat{q}$ is an alternative filler for the diagram. Since $\tilde{q}f = \hat{q}f$ , it follows that $(\tilde{q}-\hat{q})f=0$ . Since is an epimorphism, this implies that $\tilde{q}-\hat{q}=0$ . Hence we have shown that is a cokernel for . $\qedsymbol$