Kelley's proof that projection is an open map Why does Kelley (General Topology) go through the trouble of "making a copy" of the factor space in Theorem 2 of Chapter 3? The projection of an open basis set is clearly open by inspection, as Dugundji notes in his proof of the same theorem. This seems enough since the image of a union is the union of images... Thanks!

 A: It appears that Kelley constructs this map $f$ in order to have a right inverse for $P_c$. The inclusion $P_c[V]\subseteq U_c$ is obvious. The other inclusion
$$U_c\subseteq P_c[V]$$ is less trivial. For each $z\in U_c$ we have to find a point in $V$ wich projects to $z$. If we call this point $f(z)$, then this is almost equivalent to finding a right inverse $f$ to $P_c$. This $f(z)$ shall be the point $(f(z)_c)_c$ such that $f(z)_c=z$ and $f(z)_a=x_a$. These coordinates determine uniquely the point $f(z)$ (which is indeed in $V$ as $x_a\in U_a$ and $z\in U_z$) and it clearly projects to $z$, since $P_c\circ f=\text{Id}_{X_c}$.
After all, the basic idea is that an $x$ in $V$ can only exist if all $U_a$ for $a\in F$ and all $X_b$ for $b\notin F$ are non-empty. If he were less pedantic, he would just have written that for any $z\in U_a$ we can find a point in $V$ which projects to $z$, but he decided to make this point a function in $z$.
Note that $f$ is continuous and is a bijection onto its image. Since its inverse $P_c$ is also continuous when restricted to $f[X_c]$, it follows that $f$ is a homeomorphism of the factor $X_c$ to a subspace of the product space. This fact is useful, for example when you want to show that a property of the product space holds for every factor, e.g. $\prod X_a$ Hausdorff $\implies X_a$ Hausdorff, as you only need to show that it holds for subspaces.
