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Thread: Help w/ Proof in Category Theory

  1. #1
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    Help w/ Proof in Category Theory

    I don't understand this proof, specifically the part in red, I don't understand. Please help me understand this step in the proof. Thanks!

    Let Tors be the category whose objects are torsion abelian
    groups; if $\displaystyle A$ and $\displaystyle B$ are torsion abelian groups, we define $\displaystyle \text{Mor}_{\text{\textbf{Tors}}}(A, B)$ to
    be the set of all (group) homomorphisms $\displaystyle \phi : A \rightarrow B$. Prove that direct
    products exist in Tors; that is, show that given any indexed family $\displaystyle {A_i}_{i \in I}$
    where each $\displaystyle A_i $ is a torsion abelian group, there exists a torsion abelian group
    which serves as a direct product for this family in Tors.


    Proof- Let $\displaystyle T$ be the torsion subgroup (that is, the subgroup of elements of finite
    order) of $\displaystyle P = \prod_{i \in I} A_i = \{f : I \rightarrow A_i : f(i) \in A_i \text{ } \forall i\}$; here of course
    $\displaystyle P$ is the direct product of $\displaystyle \{A_i\}_{i \in I}$ in the category Ab. Let $\displaystyle j : T \rightarrow P$ be
    the inclusion and for each $\displaystyle i \in I$, let $\displaystyle \pi_i : P \rightarrow A_i$ denote the usual projection
    map; that is, $\displaystyle \pi_i(f) = f(i)$. (In coordinate notation, $\displaystyle \pi_i(a_0, a_1, \ldots) = a_i$.)
    For each $\displaystyle i \in I$ define $\displaystyle \tau_i : T \rightarrow A_i$ by $\displaystyle \tau_i = \pi_i * j$. I claim that the group
    $\displaystyle T$ together with the maps $\displaystyle \{\tau_i\}_{i \in I}$ constitute a direct product for $\displaystyle A_{i \in I}$ in
    Tors. Well, given a torsion group $\displaystyle S$ and maps $\displaystyle \sigma_i : S \rightarrow A_i$ for each $\displaystyle i \in I$,
    one defines $\displaystyle h : S \rightarrow T$ as follows: given $\displaystyle s \in S$, let $\displaystyle h(s) \in T$ be the function
    defined by $\displaystyle \color{red}{[h(s)](i) = \sigma_i(s)}$. Then clearly $\displaystyle \tau_i * h = \sigma_i \text{ } \forall i \in I$. Moreover,
    if $\displaystyle h' : S \rightarrow T$ is any other map such that $\displaystyle \tau_i * h' = \sigma_i$, then for any $\displaystyle s \in S$ and
    $\displaystyle i \in I$, $\displaystyle \color{red}{[h(s)](i) }= (\tau_i * h)(s) = \sigma_i(s) = (\tau_i * h')(s) = \color{red}{[h'(s)](i)}$, so $\displaystyle h = h'$.


    I don't understand what $\displaystyle \color{red}{[h(s)](i) = \sigma_i(s)}$ is.
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  2. #2
    Member
    Joined
    Dec 2008
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    Function-valued functions can be confusing; I've always found that stepping through the notation in painstaking detail is very helpful.

    By definition, h : S --> T
    Therefore, $\displaystyle h(s) \in T$
    By definition of T, h(s) is a function $\displaystyle I \to A_i$
    h(s)(i) is, therefore, evaluating h(s) at i. (And $\displaystyle h(s)(i) \in A_i$)
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