Proof.
(See also the remark below.) If the map f satisfies (1), it automatically satisfies (4) because any single point is quasi-compact.
Assume map f satisfies (4). We will prove it is universally closed, i.e., (3) holds. Let g : Z \to Y be a continuous map of topological spaces and consider the diagram
\xymatrix{ Z \times _ Y X \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Z \ar[r]^ g & Y }
During the proof we will use that Z \times _ Y X \to Z \times X is a homeomorphism onto its image, i.e., that we may identify Z \times _ Y X with the corresponding subset of Z \times X with the induced topology. The image of f' : Z \times _ Y X \to Z is \mathop{\mathrm{Im}}(f') = \{ z : g(z) \in f(X)\} . Because f(X) is closed, we see that \mathop{\mathrm{Im}}(f') is a closed subspace of Z. Consider a closed subset P \subset Z \times _ Y X. Let z \in Z, z \not\in f'(P). If z \not\in \mathop{\mathrm{Im}}(f'), then Z \setminus \mathop{\mathrm{Im}}(f') is an open neighbourhood which avoids f'(P). If z is in \mathop{\mathrm{Im}}(f') then (f')^{-1}\{ z\} = \{ z\} \times f^{-1}\{ g(z)\} and f^{-1}\{ g(z)\} is quasi-compact by assumption. Because P is a closed subset of Z \times _ Y X, we have a closed P' of Z \times X such that P = P' \cap Z \times _ Y X. Since (f')^{-1}\{ z\} is a subset of P^ c = P'^ c \cup (Z \times _ Y X)^ c, and since (f')^{-1}\{ z\} is disjoint from (Z \times _ Y X)^ c we see that (f')^{-1}\{ z\} is contained in P'^ c. We may apply the Tube Lemma 5.17.1 to (f')^{-1}\{ z\} = \{ z\} \times f^{-1}\{ g(z)\} \subset (P')^ c \subset Z \times X. This gives V \times U containing (f')^{-1}\{ z\} where U and V are open sets in X and Z respectively and V \times U has empty intersection with P'. Then the set V \cap g^{-1}(Y-f(U^ c)) is open in Z since f is closed, contains z, and has empty intersection with the image of P. Thus f'(P) is closed. In other words, the map f is universally closed.
The implication (3) \Rightarrow (2) is trivial. Namely, given any topological space Z consider the projection morphism g : Z \times Y \to Y. Then it is easy to see that f' is the map Z \times X \to Z \times Y, in other words that (Z \times Y) \times _ Y X = Z \times X. (This identification is a purely categorical property having nothing to do with topological spaces per se.)
Assume f satisfies (2). We will prove it satisfies (1). Note that f is closed as f can be identified with the map \{ pt\} \times X \to \{ pt\} \times Y which is assumed closed. Choose any quasi-compact subset K \subset Y. Let Z be any topological space. Because Z \times X \to Z \times Y is closed we see the map Z \times f^{-1}(K) \to Z \times K is closed (if T is closed in Z \times f^{-1}(K), write T = Z \times f^{-1}(K) \cap T' for some closed T' \subset Z \times X). Because K is quasi-compact, K \times Z\to Z is closed by Lemma 5.17.3. Hence the composition Z \times f^{-1}(K)\to Z \times K \to Z is closed and therefore f^{-1}(K) must be quasi-compact by Lemma 5.17.3 again.
\square
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