# 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 $$A$$ 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.

## Answers and Replies

Hurkyl
Staff Emeritus
Science Advisor
Gold Member
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 $h(s)(i) \in A_i$)

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 $h(s)(i) \in A_i$)

Got it, thanks.