Help w/ Proof in Category Theory

**mingcai6172** · January 6th 2009, 05:00 PM

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

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

I don't understand what $\color{red}{[h(s)](i) = \sigma_i(s)}$ is.

**GaloisTheory1** · January 6th 2009, 05:53 PM

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

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