The following is taken from An introduction to Category theory by Harold Simmons.
$\color{green}{Background}$
$\textbf{(1)}$ $\textbf{Definition:}$ Let $\nabla=\mathbb\{I,E\}$ be a directed graph viewed as a template of the first kind. Let $C$ be a category. A $\nabla-$diagram in $C$ is
an $\mathbb{I}-$indexed family of objects of $C$ $\quad$ $\mathrm{A}=\{A(i)\mid i\in\mathbb{I}\}$
an $\mathbb{}-$indexed family of arrows of $C$ $\quad$ $\mathcal{A}=\{A(e)\mid e\in\mathbb{E}\}$
Where each edge, as on the left, produces an arrow, as on the right
$$i\xrightarrow{e}j \quad\quad\quad A(i)\xrightarrow{A(e)}A(j)$$
with indicated source and target restrictions.
....As usual we have a template $\nabla$ of nodes $\mathbb{I}$ and edges
$$\mathrm{A}=(A(i)\mid i\in \mathbb{I}) \quad\quad \mathcal{A}=(A(e)\mid i\in \mathbb{E})$$
to forma a $\textbf{Top}-$diagram. Thus each $A(i)$ is a topological space and each $A(e)$ is a continuous map. By passing to $\textbf{Set}$ we obtain the set $A$ of all threads
$$\alpha:\mathbb{I}\to \cup \mathrm{A}$
together with the $\mathbb{I}-$indexed family
$$A\xrightarrow{\alpha(i)}A(i)$$
of evaluation functions. Our first job is to convert $A$ into a topological space in such a way that each $\alpha(i)$ is continuous.
Consider any node $i\in \mathbb{I}$ and any open $U\in \mathcal{O}A(i)$ of that component. We certainly require the inverse image set
$$i(U)=\alpha(i)^{\leftarrow}(U)=\{a\in A\mid a(i)\in U\}$$
to be open. Thus we take the family of all these subsets of $A$ as a subbase of a topology on $A.$ This ensures that each $\alpha(i)$ is continuous.
In the usual way this gives us a left solution of the diagram. This has nothing much to do with the topological aspects. It's merely that certain triangles of function ddo commute.
Our main job is to show that this left solution is universal in $\textbf{Top}.$ To do that we compare it with an arbitrary left solution of the $\textbf{Top}-$diagram. Thus we assume given a topological space $X$ together with an $\mathbb{I}-$indexed family of continuous maps
$$X\xrightarrow{\zeta(i)}A(i)$$
such that the $\textbf{Top}-$triangle
commutes for each edge $e,$ as indicated. We require a unique mediator
$$X\xrightarrow{\mu}A$$
which of course, must be a continuous map.
By passing to $\textbf{Set}$ the only possible mediating function is given by
$$\color{blue}{\mu(x)(i)}=\zeta(i)(x)$$
for each $x\in X$ and $i\in \mathbb{I}.$ Thus it suffices to show that this function $\mu$ is continuous. To do that it suffices to show that for each subbasic open set of $A$ the inverse image across $\mu$ is open. Thus we require
$$\mu^{\leftarrow}(i(U))\in \mathcal{O}R \text{ (there might be a misprint in notation here.)}$$
for each $i\in \mathbb{I}$ and open $U\in \mathcal{O}A(i).$ For each $x\in X$ we have
$$x\in \mu^{\leftarrow}(i(U)) \Longleftrightarrow \mu(i)\in i(U)=\alpha(i)^{\leftarrow}(U)$$
$$\quad \Longleftrightarrow \alpha(i)(\mu(x))\in U$$
$$\quad \Longleftrightarrow \mu(x)(i)\in U$$
$$\quad \Longleftrightarrow \color{brown}{\zeta(i)(r)}=\mu(x)(i)=U \Longleftrightarrow x\in \zeta(i)^{\leftarrow}(U)$$
to show that
$$\quad \Longleftrightarrow \mu^{\leftarrow}(i(U))=\zeta(i)^{\leftarrow}(U)$$
for each pair $i$ and $U.$ Since each $\zeta(i)$ is continuous this shows that each $\mu^{\leftarrow}(i(U))$ is open, for the required result.
$\color{Red}{Question:}$
In both notations, $\color{blue}{\mu(x)(i)}=\zeta(i)(x),$ $\color{brown}{\zeta(i)(r)}=\mu(x)(i)=U$ where it has $\mu(x)(i), \zeta(i)(r),$ are both $i, r$ are used to serve as subscripts. Meaning $\mu(x)(i):=\mu_i(x)$ and $\zeta(i)(r):=\zeta_r(i).$ Thank you in advance