(a) Show that $\sim$ is an equivalence relation on $B$. The set of equivalence classes is called the direct limit of the direct system $\{A_i\}$ and is denoted $\varinjlim A_i$. For the rest of the exercise let $A = \varinjlim A_i$.
(b) Let $\overline{x}$ denote the class of $x$ and define $\rho_i : A_i → A$ by $\rho_i(a)=\overline{a}$. Show that if $\rho_{ij}$ is injective for all $i,j$ then $\rho_i$ is injective for all $i$ (so that each $A_i$ can be viewed as a subset of $A$).
(c) Assume $\rho_{ij}$ are all group homomorphisms. For $a∈A_i$,$b∈A_j$ show that the operation$$\overline{a}+\overline{b}=\overline{\rho_{ik}(a)+\rho_{jk}(b)}$$ is well defined, for any $k$ such that $i,j≤k$. Show that this makes $A$ into an abelian group, and that $\rho_i$ in part (b) are group homomorphisms.
(d) Show that if $A_i$ are commutative rings with 1 and $\rho_{ij}$ are ring homomorphisms that send 1 to 1, then $A$ may likewise be given the structure of a commutative ring with 1 such that $\rho_i$ are all ring homomorphisms.
(e) Under the hypotheses in (c), prove that the direct limit has the following universal property: if $C$ is any abelian group such that for each $i∈I$ there is a homomorphism $φ_i : A_i → C$ with $φ_i = φ_j \circ \rho_{ij}$ whenever $i≤j$, then there is a unique homomorphism $φ : A → C$ such that $φ \circ \rho_i = φ_i$ for all $i$.
Proof: (a) Reflexivity: $\rho_{ii}(a)=a=\rho_{ii}(a)⇒a \sim a$. Symmetry: $a \sim b$$⇒ \rho_{ik}(a)=\rho_{jk}(b) ⇒ \rho_{jk}(b)=\rho_{ik}(a) ⇒ b \sim a$ for some $k$. Transitivity: Let $c∈A_l$. $a \sim b \land b \sim c ⇒ \rho_{ik}(a)=\rho_{jk}(b) \land \rho_{jm}(b)=\rho_{lm}(c)$ for some $k$ and $m$ with $i,j≤k$ and $j,l≤m$. Let $k,m≤n$. We have $$\rho_{in}(a)=\rho_{kn} \circ \rho_{ik}(a)=\rho_{kn} \circ \rho_{jk}(b)=\rho_{jn}(b)=$$$$\rho_{mn} \circ \rho_{jm}(b)=\rho_{mn} \circ \rho_{lm}(c)=\rho_{ln}(c)$$ so that $a \sim c$.
(b) Assume $a,b∈A_i$ and $\rho_{i}(a)=\rho_{i}(b)$, so that $\overline{a}=\overline{b} ⇒ a \sim b$. Then $\rho_{ij}(a)=\rho_{ij}(b)$ for some $j$, so that $a=b$.
(c) Start with well-definedness. Let $\overline{a'}=\overline{a}$ by $\rho_{i'n}(a')=\rho_{in}(a)$ for some $i,i'≤n$, and denote $a'∈A_{i'}$, and likewise for $b'$ with $m$ in place of $n$. It suffices to show $\rho_{i'k'}(a')+\rho_{j'k'}(b') \sim \rho_{ik}(a) + \rho_{jk}(b)$ for arbitrary applicable $k'$. Choose $k^o$ such that $n,k,k',m≤k^o$ (by choosing a value greater than or equal to $n$ and $k$, and another greater than or equal to $m$ and $k'$, and letting $k^o$ be a value greater than or equal to both). We have$$\rho_{k'k^o}(\rho_{i'k'}(a')+\rho_{j'k'}(b'))=\rho_{k'k^o} \circ \rho_{i'k'}(a')+\rho_{k'k^o} \circ \rho_{j'k'}(b')=\rho_{i'k^o}(a')+\rho_{j'k^o}(b')=$$$$\rho_{nk^o} \circ \rho_{i'n}(a')+\rho_{mk^o} \circ \rho_{j'm}(b')=\rho_{nk^o} \circ \rho_{in}(a) + \rho_{mk^o} \circ \rho_{jm}(b)=$$$$\rho_{ik^o}(a)+\rho_{jk^o}(b)=\rho_{kk^o} \circ \rho_{ik}(a)+\rho_{kk^o} \circ \rho_{jk}(b)=\rho_{kk^o}(\rho_{ik}(a)+\rho_{jk}(b))$$Due to the homomorpicity of $\rho_{ij}$, we have $\overline{e}=0$ in $A$ for any $e∈A_i$ a zero and $-\overline{a}=\overline{-a}$. Associativity, closure, and abelianness are assured by their demonstration in $A_i$. Furthermore, for $a,b∈A_i$, we have $\rho_{i}(a)+\rho_{i}(b)=\overline{a}+\overline{b}=\overline{\rho_{ii}(a)+\rho_{ii}(b)}=\rho_{i}(a+b)$.
(d) This is simply part (c) with the operation of addition replaced by multiplication, together with the evident fact that $\overline{e}=1$ in $A$ for any $e∈A_i$ an identity.
(e) Any such homomorphism $φ$ is uniquely defined by $φ(\overline{a})=φ \circ \rho_{i}(a)=φ_i(a)$, so now well definedness and homomorphicity must be demonstrated: $φ(\overline{a'})=φ_{i'}(a)=φ_n \circ \rho_{i'n}(a')=φ_n \circ \rho_{in}(a)=φ_i(a)=φ(\overline{a})$, and $φ(\overline{a}+\overline{b})=φ(\overline{\rho_{ik}(a)+\rho_{jk}(b)})=φ_k(\rho_{ik}(a)+\rho_{jk}(b))=φ_i(a)+φ_j(b)=φ(\overline{a})+φ(\overline{b})$where addition is interchangeable with multiplication.$~\square$
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