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.