Corollary 2

Let $f : X \to Y$ be a proper mapping, where $X$ is Hausdorff. Then the subspace $f(X)$ of $Y$ is Hausdorff.

By reason of Proposition 5 c) we need only consider the case where $f(X) = Y$. Then the diagonal of $Y \times Y$ is the image under $f \times f$ of the diagonal of $X$, which is closed (§ 8, no. 1, Proposition 1); $f \times f$ is proper (Proposition 4); hence the diagonal of $Y \times Y$ is closed (Proposition 1) and therefore $Y$ is Hausdorff (§ 8, no. 1, Proposition 1).