- #1
mathsss2
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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.
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.