# Square is homotopy Cartesian if horizontal maps are weak equivalences

This is probably trivial but I'm not the best with category theory.

Let $M$ be a right proper model category (that is pullbacks of weak equivalences along fibrations are weak equivalences). The claim is that a commutative square

$$\require{AMScd} \begin{CD} X_1 @>{\alpha}>> X_2 \\ @VVV @VVV \\ Y_1 @>{\beta}>> Y_2 \end{CD}$$ is homotopy Cartesian if $\alpha$ and $\beta$ are weak equivalences.

My plan was to take pullbacks to get a commutative cube and use the cogluing lemma, as it's easy to make a another parallel weak equivalence to $\alpha$ and $\beta$, but I can't see how to get that $X_1\to U\times_{Y_2}X_2$ is a weak equivalence. Here I've factored $\beta$ through $U$ using a fibration and a trivial cofibration.

Am I on the right path? If not, what is the right approach?

• It's actually true in any model category (or even any category with weak equivalences) but the proof would be different. – Zhen Lin Mar 14 '16 at 8:23

This actually is trivial. The factorization $X_2\to U\to Y_2$ with the first map a trivial cofibration and the second map a fibration does it. I don't know how to delete this, but I would be happy if someone who does did; I think it's not a valuable question.