In our class, we have already proven that:
Let $Y$ is a metric space, $X$ a topological space and $H \subseteq C(X,Y)$ equicontinuous and $\Psi$ a filter base on $H$. Suppose that $pr_x(\Psi) \to \phi(x)$ for all $x \in X$.
Then, the function $x \mapsto \phi(x)$ is continuous and $\Psi \to\phi$ in the compact open topology.
Now, it is claimed that the following is an immediate corollary of this theorem:
If $Y$ is a metric space and $H \subseteq C(X,Y)$ is equicontinous, then $(\mathcal{T}_c)_H = (\mathcal{T}_p)_H$ (the former is the compact open topology, the latter the pointwise topology (i.e. subspace topology on product)
I can't however see how this follows. We proved that $\mathcal{T}_p \subseteq \mathcal{T}_c$, and one inclusion follows from this. But the other one should follow from the theorem above.
Any ideas?
OK, following your own suggestion, let $\mathrm{id}: (H, (\mathcal{T}_p)_H) \to (H, (\mathcal{T}_c)_H)$ be defined by $\mathrm{id}(f) = f$ for all $f \in H$. The fact that trivially $\mathcal{T}_p \subseteq \mathcal{T}_c$ on $Y^X$ and hence on all subspaces too (singletons are compact), $\mathrm{id}$ is an open map.
The facts you have been given show its continuity: let $f \in H$ and suppose that for some filterbase on $H$, $\Psi \to f$ pointwise (i.e. in $(\mathcal{T}_p)_H$), which means that $\pi_x[\Psi] \to \pi_x(f) = f(x)$ for all $x$. As $\Psi \to f$ in $\mathcal{T}_c$ in that case (I don't use the first fact as $f \in H$ already means that $f$ is continuous), $\mathrm{id}[\Psi] \to \textrm{id}(f)$ in $(H, (\mathcal{T}_c)_H)$, and as continuity is determined by convergence of filters or filter bases, $\mathrm{id}$ is continuous, which immediately implies the identity of topologies.
We could have extended it too $K = \overline{H}$, the pointwise closure of $H$, which is still equicontinuous, and we'd have shown that $K$ is also the compact-open closure of $H$ this way.