When I started to study topology of compact convergence it seemed to me that there was something not said explicitly that really help to understand better this topology so that I observed that for any metri space $(Y,d)$ and for any topological space $(X,\cal T)$ if $Y^X$ is equipped with the compact topology $\mathcal T\_C$ corresponding to $d$ then any projection $\pi\_K$ of $Y^X$ onto $Y^K$ with $K\in\mathcal P(X)$ compact is continuous when $Y^K$ is equipped with topology of uniform convergence $\mathcal T\_\infty(Y^K)$ and really this is true since actually $\mathcal T\_C$ is the initial topology corresponding to these projections as [here](https://math.stackexchange.com/a/4612060/736008) is proved. Anyway, when Munkres into his Topology text started to explain the topology of compact convergence then he gives as exercise to show that a sequence of functions $(f\_n)\_{n\in\omega}$ in $Y^X$ converges to any function $f$ if and only if for any compact $K\in\mathcal P(X)$ the restricted sequence $\big(f\_n|\_K\big)\_{n\in\omega}$ converges to $f|\_K$ so that I observed that this was trivial if the following proposition

> 
> **Proposition 1**
> 
> 
> If $\Sigma$ is a set equipped with the initial topology $\cal T\_\Phi$ corresponding to the collection of functions $\Phi$ then a net $(\sigma\_i)\_{i\in I}$ in $\Sigma$ based on the directed set $(I,\preceq\_I)$ converges to an $\sigma\in\Sigma$ if and only if for any $\varphi\in\Phi$ the net $\big(\varphi(\sigma\_i)\big)\_{i\in I}$ converges to $\varphi(\sigma)$
> 
> 
> 

holds.

So, I observed that if $(\sigma\_i)\_{i\in I}$ converges to any $\sigma$ then by continuity of $\varphi\in\Phi$ the net $\big(\varphi(\sigma\_i)\big)\_{i\in I}$ converges to $\varphi(\sigma)$. However, for any $\varphi\in\Phi$ let be $\Lambda\_\varphi$ the codomain of $\varphi$ and so $\cal T\_\varphi$ the corresponding topology for $\Lambda\_\varphi$ so that if $A\_\sigma\in\cal T\_\Phi$ is a(n) (open) neighborhood of $\sigma$ then there exists $(\boldsymbol\varphi,\boldsymbol A)\in\prod\_{p\in q}\Phi\times\prod\_{p\in q}\cal T\_{\varphi\_p}$ with $q\in\omega$ such that
$$
\sigma\in\bigcap\_{p\in q}\varphi^{-1}\_p[A\_p]\subseteq A\_\sigma
$$
where we put
$$
\varphi\_p:=\boldsymbol\varphi(p)\quad\text{and}\quad A\_p:=\boldsymbol A(p)
$$
for any $p\in q$. So if $\big(\varphi(\sigma\_i)\big)\_{i\in I}$ converges to $\varphi(\sigma)$ for each $\varphi\in\Phi$ then for any $p\in q$ there exists $i\_p\in I$ such that
$$
\varphi(\sigma\_i)\in A\_p
$$
for any $i\succeq\_I i\_p$ so that fixing **by cofinality** $i\_0\in I$ such that
$$
i\_p\preceq\_I i\_0
$$
for any $p\in q$ then for any $i\succeq\_I i\_0$ we have
$$
\sigma\_i\in\bigcap\_{p\in q}\varphi^{-1}\_p[A\_p]\subseteq A\_\sigma
$$
which proves that $(\sigma\_i)\_{i\in I}$ converges to $\sigma$.

So, I realise that what I did can appear a not hard result so that the question can be reputed unnecessary but I decided to pose it because many (notable) authors (e.g. Engelking, Willard, Kelley etc...) prove proposition $1$ only when $\Sigma$ is just a product of any collection $\frak S$ and this led me to think that it could not be true if this is not. So, could someone help me, please?