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 thus asking to me if $\mathcal T_C$ is actually the initial topology corresponding to these projections I observed what to follows.
Let be $(Y,d)$ a metric space and thus if $\mathcal T_X$ is a topology on a set $X$ then for any $K\in\mathcal P(X)$ compact with respect $\mathcal T_X$, for any $\epsilon\in\Bbb R^+$ and for any $v\in Y^X$ we can put
$$
B_K(v,\epsilon):=\Big\{u\in Y^X:\sup_{x\in K} d\big(v(x),u(x)\big)<\epsilon\Big\}
$$
so that if $\cal K$ is the set of all compact subspace of $X$ with respect $\mathcal T_X$ then putting
$$
\mathcal B_v:=\big\{B_K(v,\epsilon):(K,\epsilon)\in\cal K\times\Bbb R^+\big\}
$$
for any $v\in Y^X$ the collection
$$
\mathcal B:=\bigcup_{v\in Y^X} \mathcal B_v
$$
is a base for a topology $\mathcal T_C$ on $Y^X$, known as topology of compact convergence.
Now so that let's we prove that for any compact $K$ the projection
$$
\pi_K:Y^X\ni y_X\longrightarrow y_X|_K\in Y^K
$$
is continuous when $Y^X$ and $Y^K$ are respectively equipped topology of compact convergence and with topology of uniform convergence.
So fixed $v_X\in Y^X$ we observe that for any $\epsilon\in\Bbb R^+$ the equality
$$
\begin{equation}
\tag{1}\label{1}
\pi_K^{-1}\Big[B\big(\pi_K(v_X),\epsilon\big)\Big]=
\Big\{u_X\in Y^X:\pi_K(u_X)\in B\big(\pi_K(v_X),\epsilon\big)\Big\}\\ =
\Big\{u_X\in Y^X:u_X|_K\in B(v_X|_K,\epsilon)\Big\} \\ =
\Big\{u_X\in Y^X:\sup_{x\in K}d\big(v_X|_K(x),u_X|_K(x)\big)<\epsilon\Big\} \\ =
\Big\{u_X\in Y^X:\sup_{x\in K}d\big(v_X(x),u_X(x)\big)<\epsilon\Big\}=
B_K(v_X,\epsilon)
\end{equation}
$$
holds and thus continuity of $\pi_K$ follows immediately since it proves that the primage of a basic open neighborhood of $\pi_K(y_X)$ is a (basic) open neighborhood of $y_X$.
So by continuity of $\pi_K$ with $K\in\cal K$ and by \eqref{1} we conclude that the collection
$$
\mathcal S:=\big\{\pi^{-1}_K[A_K]:(A_K,K)\in\mathcal T_\infty(Y^K)\times\mathcal K\big\}
$$
is a subbase for $\mathcal T_C$ since by continuity $\cal S$ is an "open" collection and since by \eqref{1} it contains $\cal B$ which is a base: this means that $\mathcal T_C$ is actually the initial topology corresponding to the "compact" projection of $Y^X$ onto $Y^K$ equipped with uniform topology where $K\in\mathcal P(X)$ is compact.
So I ask if actually the topology of compact convergence is the initial topology corresponding to "compact" projections: could someone help me, please?
Best Answer
For a compact $K$, the set $Y^K$ can be endowed with the $\sup$-metric $d_\infty(v,w) = \sup\{ d(v(x),w(x)) \mid x \in K \}$. The topology of uniform convergence $\mathcal T_∞(Y^K)$ is nothing else than the metric topology with respect to $d_\infty$. A base $\mathcal B_K$ for this topology is given by the set all open metric balls $$U(K,w,\epsilon) = \{w' \in Y^K \mid d_\infty(w',w) < \epsilon \}$$ with $w \in Y^K$ and $\epsilon \in \mathbb R^+$.
The functions $\pi_K : Y^X \to (Y^K,\mathcal T_∞(Y^K))$ induce the initial topology $\mathcal T_{init}$ on the set $Y^X$ by taking as a subbase $$\mathcal S_{init} = \bigcup_{K \in \mathcal K} \pi_K^{-1}(\mathcal T_∞(Y^K)),$$ where $ \pi_K^{-1}(\mathcal T_∞(Y^K)) = \{\pi_K^{-1}(U) \mid U \in \mathcal T_∞(Y^K))\}$. But then $$S'_{init} = \bigcup_{K \in \mathcal K} \pi_K^{-1}(\mathcal B_K)$$ is also a subbase for $\mathcal T_{init}$. This is true because each element of $\mathcal S_{init}$ is a union of elements of $S'_{init}$.
In fact we have $B_K(v,\epsilon) = \pi_K^{-1}(U(K,v\mid_K,\epsilon))$ because $v' \in B_K(v,\epsilon)$ iff $\lVert v'\mid_K - v \mid_K \rVert < \epsilon$ iff $\pi_K(v') \in U(K,v \mid_K,\epsilon)$ iff $v' \in \pi_K^{-1}(U(K,v\mid_K,\epsilon))$.
Note that it is not guaranteed that $\pi_K^{-1}(U(K,w,\epsilon)) \in \mathcal B$. By 1. this would be true if all $w \in Y^K$ have the form $w = v \mid_K$ for some $v \in Y^X$.
To show that each $\pi_K^{-1}(U(K,w,\epsilon)) \in \mathcal T_C$, consider $v \in \pi_K^{-1}(U(K,w,\epsilon))$. This means that $v \mid_K = \pi_K(v) \in U(K,w,\epsilon)$, i.e. we get $\lVert v \mid_K - w \rVert_\infty < \epsilon$. Thus $\delta = \epsilon - \lVert v \mid_K - w \rVert_\infty > 0$. We claim that $B_K(v,\delta) \subset \pi_K^{-1}(U(K,w,\epsilon))$ (this allows to conclude that $\pi_K^{-1}(U(K,w,\epsilon)) \in \mathcal T_C$).
For $v' \in B_K(v,\delta)$ we have $\lVert v'\mid_K - v \mid_K \rVert_\infty < \delta$ and therefore $\lVert v' \mid_K - w \rVert_\infty \le \lVert v'\mid_K - v \mid_K \rVert_\infty + \lVert v \mid_K - w \rVert_\infty < \delta + \lVert v \mid_K - w \rVert_\infty = \epsilon$, i.e. $\pi_K(v') = v' \mid_K \in U(K,w,\epsilon)$. Hence $v' \in \pi_K^{-1}(U(K,w,\epsilon))$.
Clearly 1. and 2. prove that $\mathcal T_{init} = \mathcal T_C$.
Let me remark that one can easily show that $\mathcal T_C$ is nothing else than the compact-open topology on $Y^X$.