Help w/ Proof in Category Theory

  • Thread starter mathsss2
  • Start date
  • #1
38
0
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 [tex]A[/tex] and [tex]B[/tex] are torsion abelian groups, we define [tex]\text{Mor}_{\text{\textbf{Tors}}}(A, B)[/tex] to
be the set of all (group) homomorphisms [tex]\phi : A \rightarrow B[/tex]. Prove that direct
products exist in Tors; that is, show that given any indexed family [tex]{A_i}_{i \in I}[/tex]
where each [tex]A_i [/tex] is a torsion abelian group, there exists a torsion abelian group
which serves as a direct product for this family in Tors.


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


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

Answers and Replies

  • #2
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
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, [itex]h(s) \in T[/itex]
By definition of T, h(s) is a function [itex]I \to A_i[/itex]
h(s)(i) is, therefore, evaluating h(s) at i. (And [itex]h(s)(i) \in A_i[/itex])
 
  • #3
38
0
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, [itex]h(s) \in T[/itex]
By definition of T, h(s) is a function [itex]I \to A_i[/itex]
h(s)(i) is, therefore, evaluating h(s) at i. (And [itex]h(s)(i) \in A_i[/itex])

Got it, thanks.
 

Related Threads on Help w/ Proof in Category Theory

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
1K
Replies
4
Views
4K
Replies
2
Views
1K
Replies
8
Views
1K
Replies
7
Views
11K
  • Last Post
Replies
13
Views
4K
Replies
11
Views
3K
Top