Proposition 6

Let $X$ be a Souslin space, let $Y$ be a Hausdorff space and let $f : X \to Y$ be a continuous mapping. Then the inverse image under $f$ of a Souslin subspace $A$ of $Y$ is a Souslin subspace of $X$.

Let $P, Q$ be Polish spaces, let $g$ be a continuous mapping of $P$ onto $X$ and let $h$ be a continuous mapping of $Q$ onto $A$. Let $R$ be the set of all $(x, y) \in P \times Q$ such that $f(g(x)) = h(y)$; $R$ is closed in $P \times Q$ and is therefore a Polish subspace of $P \times Q$ (no. 1, Proposition 1). Let $\varphi$ be the restriction to $R$ of the projection $pr_1$. Then the subspace $\overline{f}^{-1}(A)$ of $X$ is the image of $R$ under the continuous mapping $g \circ \varphi$ and is therefore a Souslin space.