The following is taken from "Elementary Topology and Applications" by: Carlos R. Borges, "Foundations of Topology by: C.Wayne Patty, and "Point set topology An experience of a Teacher by: Dinesh J. Karia.
$\color{green}{Background}$
[From Borges]
$\textbf{(1)}$ $\textbf{Definition 1:}$ $\textbf{General Cartesian Products}$
Let $a=\{A_\alpha\}_{\alpha\in \Lambda}$ be a family of sets. The $\textit{cartesian product}$ of the family $\alpha=\{A_\alpha\}_{\alpha\in \Lambda}$ is the set
$$\{f:\Lambda\to \cup_{\alpha\in \Lambda}A_\alpha\mid f(\beta)\in A_\beta \text{ for each }\beta\in \Lambda\}$$
and is denoted by $\prod_{\alpha\in \Lambda}A_\alpha.$ (Note that, to be precise, we should emphasize that $\prod_{\alpha\in \Lambda}A_\alpha\subset p(\Lambda\times \cup_{\alpha\in \Lambda}A_\alpha)$.) if $a$ is finite, it is easy to prove, from the aforementioned axioms of set theory, that $\prod_{\alpha\in \Lambda}A_\alpha$ is nonempty. $\textit{If it is not finite, it has been proved that, from the aforementioned axioms of set theory, one can neither deduce that}$ $\prod_{\alpha\in \Lambda}A_\alpha$ $\textit{is empty nor that it is nonempty. (We need more axioms!)}$
Given the sets $A_1,\ldots,A_n,$ with $n$ a positive integer, and letting $S=\{1,2,\ldots,n\},$ it is customary to let
$$\prod_{i=1}^{n}A_i\equiv\prod_{i\in S}A_i.$$
Also, extending the notion of an ordered pair to the notion of an "ordered n-tuple" $(a_1,\ldots,a_n),$ in some convenient way, it is customary to let
$$\prod_{i=1}^{n}A_i=\{(\mathrm{a}_1,\ldots,\mathrm{a}_n)\mid \mathrm{a}_i\in A_1,\ldots,\mathrm{a}_n\in A_n\}$$
it being clearly understood that $(a_1,\dots,a_n)$ corresponds, in a one-to-one fashion, to the function $f:S\to \cup_{i=1}^{n}A_i$ such that $f(i)=\mathrm{a}_i\in A_i,$ for $i=1,\ldots,n.$ The reader is well-advised to think in terms of functions rather than $n-$tuples, since functions impose no limitations on the $\textit{size}$ of the index set $\Lambda$: on the other hand, to think of tuples which have more elements than the integers may cause headaches.
In some instances, it is very convenient to replace $f:\Lambda\to \cup_{\alpha\in \Lambda}A_\alpha$ by its image $f(\Lambda)$ and to give it a more familiar appearance; namely, we let
$$f\equiv f(\Lambda)\equiv (f(\alpha))_\alpha\equiv (a_\alpha)_\alpha.$$
it being understood that $(a_\alpha)_\alpha$ represents the function $f:A\to \cup_{\alpha\in \Lambda}A_\alpha$ such that $f(\alpha)=a_\alpha,$ for each $\alpha\in \Lambda.$ By no means, under any circumstances, try to attach some order to $(a_\alpha)_\alpha$ since $\textit{none is implied}.$ Of course, in case $\Lambda=\omega,$ we can experience the sensation of order, by letting
$$(a_n)_n\equiv (a_0,a_1,\ldots).$$
For each family $\{A_\alpha\}_{\alpha\in \Lambda}$ of sets and $\beta\in \Lambda,$ we define the $\beta-\textit{projection}\prod_\beta:\prod_{\alpha\in \Lambda}A_\alpha\to A_\beta$ by letting $\prod_\beta f(f)=f(\beta)\in A_\beta,$ for each $f\in \prod_{\alpha\in \Lambda}A_\alpha.$ (Clearly, each $\prod_\beta$ is a function (!).)
[From Patty]
$\textbf{(2)}$ $\textbf{Definition 2:}$ Let $\{(X_\alpha, \mathcal{T}_\alpha):\alpha \in \Lambda\}$ be an indexed family of topological spaces. Let $\mathcal{B}$ be the collection of all sets of the form $\prod_{\alpha\in \alpha}U_\alpha$ where $U_\alpha \in \mathcal{T}_\alpha$ for each $\alpha\in \Lambda.$ Then $\mathcal{B}$ is a basis for a topology on $\prod_{\alpha \in \Lambda}X_\alpha,$ and this topology is called the $\textbf{box topology}.$
$\textbf{(3)}$ $\textbf{Definition 3:}$ Let $\{X_\alpha:\alpha \in \Lambda\}$ be an indexed family of sets and let $\beta\in \Lambda.$ The $\textbf{projection mapping}$ associated with $\beta$ is the function $\pi_\beta:\prod_{\alpha\in \Lambda}X_\alpha\to X_\beta$ defined by $\pi_{\beta}(\langle X_\alpha \rangle_{\alpha\in \Lambda})=x_\beta.$
$\textbf{(4)}$ $\textbf{Definition 4:}$ Let $\{(X_\alpha, \mathcal{T}_\alpha):\alpha \in \Lambda\}$ be an indexed family of topological spaces. For each $\alpha\in \Lambda,$ let $\mathcal{L}_\alpha=\{\pi^{-1}(U_\alpha):U_\alpha\in \mathcal{T}_\alpha\}$ and let $\mathcal{L}=\cup_{\alpha\in \Lambda}\mathcal{L}_\alpha.$ Then $\mathcal{L}$ is a subbasis for a topology $\mathcal{T}$ on $\prod_{\alpha\in \Lambda}X_\alpha,$ and $\mathcal{T}$ is called the $\textbf{product topology.}$ The topological space $(\prod_{\alpha\in \Lambda} X_\alpha, \mathcal{T})$ is called the $\textbf{product space}.$
[From: Karia]
$\textbf{(5)}$ $\textbf{Notation:}$ Let $X=\coprod_{i\in I}X_i$ and for $j\in I, l_j:x\in X_j\mapsto (x,j)\in X$ will be the injection maps.
$\textbf{Definition 5:}$ The strongest topology on $X$ generated by all injection maps is call $\textbf{the coproduct topology}$ on $X$ and is denoted by $\mathcal{T}_\text{cp}.$
Remark:
$(1)$ Since the topology on $X$ is the strongest topology such that each injection is continuous, by definition injections are continuous.
$(2)$ $G\subset X$ is open if and only if $G_i=\{x\in X_i: (x,i)\in G\}$ is open in $X_i$ for every $i\in I$ if and only if $G=\coprod_{i\in I}G_i,$ where $G_i$ is open in $X_i$ for each $i.$
$(3)$ $l_i(X_i)={x^{*}}_i$ and in fact, each $l_i$ is an embedding, giving a homeomorphism $l_i:X_i\to {x^{*}}_i.$
$\color{Red}{Question:}$
From the above definitions, I can tell that in Topology, the definition of the product topology is related to the definition of generalised cartesian products and also limiting the open sets that it can have. While the box-topology is the topology that has the most open sets. But for the notion of Co-product as a concept, is it a purely a categorical in nature?
Thank you in advance.