Proposition 4

Let $X$ and $Y$ be two topological spaces and $f$ a continuous mapping from $X$ to $Y$. For every irreducible subset $E$ of $X$, $f(E)$ is an irreducible subset of $Y$.

If $U, V$ are two open sets of $Y$ which meet $f(E)$, $f^{-1}(U)$ and $f^{-1}(V)$ are open sets of $X$ which meet $E$. Consequently, $f^{-1}(U) \cap f^{-1}(V) = f^{-1}(U \cap V)$ meets $E$, which implies that $U \cap V$ meets $f(E)$ and proves the proposition.