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| \PMlinkescapeword{sum} |
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| \PMlinkescapeword{satisfies} |
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| \PMlinkescapeword{similar} |
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| \PMlinkescapeword{terminal} |
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|
| \begin{claim*} |
\begin{claim*} |
| The category $\Ab$ of abelian groups is an abelian category. |
The category $\Ab$ of abelian groups is an abelian category. |
| \end{claim*} |
\end{claim*} |
|
|
| \begin{proof} |
\begin{proof} |
| We will justify all the axioms. |
We will justify all the axioms. |
|
|
| (Axiom 1.) Suppose $A$ and $B$ are abelian groups. We need to show that $\Hom(A,B)$ has the structure of an abelian group. Suppose $f\colon A\to B$ and $g\colon A\to B$ are elements of $\Hom(A,B)$, and define their sum $f+g\colon A\to B$ by the rule |
(Axiom 1.) Suppose $A$ and $B$ are abelian groups. We need to show that $\Hom(A,B)$ has the structure of an abelian group. Suppose $f\colon A\to B$ and $g\colon A\to B$ are elements of $\Hom(A,B)$, and define their sum $f+g\colon A\to B$ by the rule |
| \[ |
\[ |
| (f+g)(x) = f(x) + g(x) |
(f+g)(x) = f(x) + g(x) |
| \] |
\] |
| for any $x\in A$. This operation inherits the commutativity and associativity of the addition in $B$. Moreover, the function $f+g$ is a homomorphism. To see this, suppose $x$ and $y$ are in $A$. Then |
for any $x\in A$. This operation inherits the commutativity and associativity of the addition in $B$. Moreover, the function $f+g$ is a homomorphism. To see this, suppose $x$ and $y$ are in $A$. Then |
| \begin{align*} |
\begin{align*} |
| (f+g)(x+y) &= f(x+y) + g(x+y) = f(x) + g(x) + f(y) + g(y) \\ |
(f+g)(x+y) &= f(x+y) + g(x+y) = f(x) + g(x) + f(y) + g(y) \\ |
| &= (f+g)(x) + (f+g)(y). |
&= (f+g)(x) + (f+g)(y). |
| \end{align*} |
\end{align*} |
| The identity in $\Hom(A,B)$ is the constant zero function $0\colon A\to B$, since for any $x\in A$, |
The identity in $\Hom(A,B)$ is the constant zero function $0\colon A\to B$, since for any $x\in A$, |
| \[ |
\[ |
| (f+0)(x) = f(x) + 0(x) = f(x) + 0 = f(x). |
(f+0)(x) = f(x) + 0(x) = f(x) + 0 = f(x). |
| \] |
\] |
| Thus $\Hom(A,B)$ is an abelian group. |
Thus $\Hom(A,B)$ is an abelian group. |
|
|
| Now we show that composition of morphisms distributes over addition in $\Hom(\cdot,\cdot)$. Suppose we are given a diagram |
Now we show that composition of morphisms distributes over addition in $\Hom(\cdot,\cdot)$. Suppose we are given a diagram |
| \[\xymatrix{ |
\[\xymatrix{ |
| A \ar[r]^f & B\ar@/^/[r]^g\ar@/_/[r]_h & C\ar[r]^k & D |
A \ar[r]^f & B\ar@/^/[r]^g\ar@/_/[r]_h & C\ar[r]^k & D |
| }\] |
}\] |
| of abelian groups. We claim that $(g+h)f = gf+hf$ and $k(g+h) = kg+kh$. Since the proofs are similar, we prove only the first identity. Let $x\in A$. Then |
of abelian groups. We claim that $(g+h)f = gf+hf$ and $k(g+h) = kg+kh$. Since the proofs are similar, we prove only the first identity. Let $x\in A$. Then |
| \begin{align*} |
\begin{align*} |
| ((g+h)f)(x) &= (g+h)(f(x)) = g(f(x)) + h(f(x)) \\ |
((g+h)f)(x) &= (g+h)(f(x)) = g(f(x)) + h(f(x)) \\ |
| &= (gf)(x) + (hf)(x) = (gf+hf)(x). |
&= (gf)(x) + (hf)(x) = (gf+hf)(x). |
| \end{align*} |
\end{align*} |
| Thus $\Ab$ satisfies Axiom 1. |
Thus $\Ab$ satisfies Axiom 1. |
|
|
| (Axiom 2.) The trivial group $0$ is a zero object in $\Ab$. It is initial because there exists a unique morphism $0\to A$ for any abelian group $A$, and it is terminal because there exists a unique morphism $A\to 0$ for any abelian group $A$. In both cases the morphism is the constant zero function. |
(Axiom 2.) The trivial group $0$ is a zero object in $\Ab$. It is initial because there exists a unique morphism $0\to A$ for any abelian group $A$, and it is terminal because there exists a unique morphism $A\to 0$ for any abelian group $A$. In both cases the morphism is the constant zero function. |
|
|
| (Axiom 3.) The \PMlinkname{Cartesian product}{DirectProductAndRestrictedDirectProductOfGroups} of two abelian groups is a categorical direct product. Since the Cartesian product of two abelian groups is again abelian, it follows that $\Ab$ has finite products. |
(Axiom 3.) The \PMlinkname{Cartesian product}{DirectProductAndRestrictedDirectProductOfGroups} of two abelian groups is a categorical direct product. Since the Cartesian product of two abelian groups is again abelian, it follows that $\Ab$ has finite products. |
|
|
| (Axiom 4.) Now we show that $\Ab$ 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 $K$ is the \PMlinkname{group-theoretic kernel}{KernelOfAGroupHomomorphism} of $f$, hence a normal subgroup of $A$. Since $A$ is abelian, so is $K$. The inclusion $i\colon K\to A$ is a morphism in $\Ab$. Moreover, by construction we have that $fi=0$. |
(Axiom 4.) Now we show that $\Ab$ 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 $K$ is the \PMlinkname{group-theoretic kernel}{KernelOfAGroupHomomorphism} of $f$, hence a normal subgroup of $A$. Since $A$ is abelian, so is $K$. The inclusion $i\colon K\to A$ is a morphism in $\Ab$. Moreover, by construction we have that $fi=0$. |
|
|
| So suppose that $j\colon L\to A$ is a morphism in $\Ab$ such that $fj=0$. We need to show that the diagram |
So suppose that $j\colon L\to A$ is a morphism in $\Ab$ such that $fj=0$. We need to show that the diagram |
| \[\xymatrix{ |
\[\xymatrix{ |
| & L\ar[d]^j\ar@{.>}[dl]_{\tilde{\j}} & \\ |
& L\ar[d]^j\ar@{.>}[dl]_{\tilde{\j}} & \\ |
| K\ar[r]_i & A\ar[r]_f & B |
K\ar[r]_i & A\ar[r]_f & B |
| }\] |
}\] |
| has a unique filler $\tilde{\j}$. Let $x\in L$. Since $fj(x)=0$, 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 $\Ab$ has kernels. |
has a unique filler $\tilde{\j}$. Let $x\in L$. Since $fj(x)=0$, 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 $\Ab$ has kernels. |
|
|
| Now we construct a cokernel for $f$. Define a subset $I\subset B$ by |
Now we construct a cokernel for $f$. Define a subset $I\subset B$ by |
| \[ |
\[ |
| I = \{ f(x) \colon x\in A\}. |
I = \{ f(x) \colon x\in A\}. |
| \] |
\] |
|
Then $I$ is a subgroup of the abelian subgroup $B$, so we may form the quotient group $C=B/I$. Thus $C$ is the group-theoretic cokernel of $f$. Define a group homoomorphism $p\colon B\to C$ by $p(x) = x + I$. Since $C$ is a quotient of an abelian group, it is abelian, so $p\colon B\to C$ is a morphism in $\Ab$. Moreover, by construction we have that $pf = 0$.
|
Then $I$ is a subgroup of the abelian subgroup $B$, so we may form the quotient group $C=B/I$. Thus $C$ is the group-theoretic cokernel of $f$. Define a group homoomorphism $p\colon B\to C$ by $p(x) = x + I$. Since $C$ is a quotient of an abelian group, it is abelian, so $p\colon B\to C$ is a morphism in $\Ab$. Moreover, by construction we have that $pf = 0$.
|
|
|
| So suppose that $q\colon A\to D$ is a morphism in $\Ab$ such that $qf=0$. We need to show that the diagram |
So suppose that $q\colon A\to D$ is a morphism in $\Ab$ such that $qf=0$. We need to show that the diagram |
| \[\xymatrix{ |
\[\xymatrix{ |
| A\ar[r]^f & B\ar[r]^p\ar[d]_q & C\ar@{.>}[dl]^{\tilde{q}} \\ |
A\ar[r]^f & B\ar[r]^p\ar[d]_q & C\ar@{.>}[dl]^{\tilde{q}} \\ |
| & D |
& D |
| }\] |
}\] |
| has a unique filler. Let $x+I\in C$. Suppose $y$ is another representative in $B$ for $x+I$, that is, that $p(x) = p(y)$. Then $x-y\in I$, that is, there is some $w\in A$ such that $f(w) = x - y$. By assumption, $qf = 0$, so $q(x) = q(y)$. So there is a well-defined function $\tilde{q}\colon C\to D$ defined by $\tilde{q}(x+I) = q(x)$. Moreover, for any $x+I$ and $y+I$ in $C$, |
has a unique filler. Let $x+I\in C$. Suppose $y$ is another representative in $B$ for $x+I$, that is, that $p(x) = p(y)$. Then $x-y\in I$, that is, there is some $w\in A$ such that $f(w) = x - y$. By assumption, $qf = 0$, so $q(x) = q(y)$. So there is a well-defined function $\tilde{q}\colon C\to D$ defined by $\tilde{q}(x+I) = q(x)$. Moreover, for any $x+I$ and $y+I$ in $C$, |
| \[ |
\[ |
| \tilde{q}(x+y+I) = q(x+y) = q(x) + q(y) = \tilde{q}(x+I) +\tilde{q}(y+I), |
\tilde{q}(x+y+I) = q(x+y) = q(x) + q(y) = \tilde{q}(x+I) +\tilde{q}(y+I), |
| \] |
\] |
| so $\tilde{q}$ is a morphism in $\Ab$. 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 $\Ab$ has cokernels. |
so $\tilde{q}$ is a morphism in $\Ab$. 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 $\Ab$ has cokernels. |
|
|
| (Axiom 5.) Suppose $f\colon A\to B$ is a monomorphism, and let $p\colon B\to\coker f$ be a cokernel of $f$. We must show that $f$ is a kernel of $p$. Since $p$ is a cokernel of $f$, we know that $pf = 0$. Now we must show that if $j\colon L\to B$ is any morphism in $\Ab$ such that $pj = 0$, then the diagram |
(Axiom 5.) Suppose $f\colon A\to B$ is a monomorphism, and let $p\colon B\to\coker f$ be a cokernel of $f$. We must show that $f$ is a kernel of $p$. Since $p$ is a cokernel of $f$, we know that $pf = 0$. Now we must show that if $j\colon L\to B$ is any morphism in $\Ab$ such that $pj = 0$, then the diagram |
| \[\xymatrix{ |
\[\xymatrix{ |
| & L\ar[d]^j\ar@{.>}[dl]_{\tilde{\j}} & \\ |
& L\ar[d]^j\ar@{.>}[dl]_{\tilde{\j}} & \\ |
| A\ar[r]_f & B\ar[r]_<<<p & \coker f |
A\ar[r]_f & B\ar[r]_<<<p & \coker f |
| }\] |
}\] |
| has a unique filler $\tilde{\j}$. So suppose $x\in L$. Since $pj(x) = 0$, it follows by the construction of the cokernel given above that there is some $y\in A$ such that $f(y) = j(x)$. Since $f$ is monomorphism, $f$ is injective, so this $y$ is unique. We may therefore define $\tilde{\j}$ by the formula $\tilde{\j}(x) = f^{-1}(j(x))$. Hence $f$ is a kernel for the cokernel of $f$. |
has a unique filler $\tilde{\j}$. So suppose $x\in L$. Since $pj(x) = 0$, it follows by the construction of the cokernel given above that there is some $y\in A$ such that $f(y) = j(x)$. Since $f$ is monomorphism, $f$ is injective, so this $y$ is unique. We may therefore define $\tilde{\j}$ by the formula $\tilde{\j}(x) = f^{-1}(j(x))$. Hence $f$ is a kernel for the cokernel of $f$. |
|
|
| (Axiom 6.) Suppose $f\colon A\to B$ is an epimorphism, and let $i\colon \ker f\to A$ be a kernel of $f$. We must show that $f$ is a cokernel of $i$. Since $i$ is a kernel of $f$, we know that $fi = 0$. Now we must show that if $q\colon A\to D$ is any morphism in $\Ab$ such that $qi=0$, then the diagram |
(Axiom 6.) Suppose $f\colon A\to B$ is an epimorphism, and let $i\colon \ker f\to A$ be a kernel of $f$. We must show that $f$ is a cokernel of $i$. Since $i$ is a kernel of $f$, we know that $fi = 0$. Now we must show that if $q\colon A\to D$ is any morphism in $\Ab$ such that $qi=0$, then the diagram |
| \[\xymatrix{ |
\[\xymatrix{ |
| \ker f\ar[r]^i & A\ar[r]^f\ar[d]_q & B\ar@{.>}[dl]^{\tilde{q}} \\ |
\ker f\ar[r]^i & A\ar[r]^f\ar[d]_q & B\ar@{.>}[dl]^{\tilde{q}} \\ |
| & D & |
& D & |
| }\] |
}\] |
| has a unique filler $\tilde{q}$. |
has a unique filler $\tilde{q}$. |
|
|
| To prove the existence of $\tilde{q}$, first recall that epimorphisms in $\Ab$ are surjections. Let $z\in B$. Suppose $f(x)=f(y)=0$. Then $x-y$ is in $\ker f$. By assumption, $qi=0$, so this implies that $q(x)=q(y)$. So we may define a morphism $\tilde{q}\colon B\to D$ in $\Ab$ by the formula $\tilde{q}(z) = q(x)$, where $x$ is any element of $f^{-1}(z)$. Moreover, $\tilde{q}f = q$ by construction. |
To prove the existence of $\tilde{q}$, first recall that epimorphisms in $\Ab$ are surjections. Let $z\in B$. Suppose $f(x)=f(y)=0$. Then $x-y$ is in $\ker f$. By assumption, $qi=0$, so this implies that $q(x)=q(y)$. So we may define a morphism $\tilde{q}\colon B\to D$ in $\Ab$ by the formula $\tilde{q}(z) = q(x)$, where $x$ 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 $f$ is an epimorphism, this implies that $\tilde{q}-\hat{q}=0$. Hence we have shown that $f$ is a cokernel for $i$. |
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 $f$ is an epimorphism, this implies that $\tilde{q}-\hat{q}=0$. Hence we have shown that $f$ is a cokernel for $i$. |
| \end{proof} |
\end{proof} |
