Let $\mathcal{P}$ be a property of morphisms of schemes that is fulfilled by isomorphisms.
Now assume we have a property $\mathcal{P}$ that is invariant under base-change, that means if $f:X \to Y$ satisfies $\mathcal{P}$, then for any morphism $g:Y' \to Y$ so does $f_{y'}:X_{Y'} \to Y'$.
Then why do we also have that if the diagonal morphism $\Delta_{X/Y}:X \to X \times_{Y} X$ satisfies $\mathcal{P}$, then so does the diagonal morphism after base-change, so $\Delta_{X_{Y'} / Y'}$ ?
Our lecturer said, this was clear. But I am really struggling to use the concepts of base-change and the diagonal. So any hint would be appreciated.
My ideas:
In the beginning we have the commutative diagram:
$$\require{AMScd}$$ \begin{CD} X\times_{Y}X @>{p_{1}}>> X\\ @VV{p_{2}}V @VV{f}V\\ X @>{f}>> Y \end{CD}
as well as
$$\require{AMScd}$$ \begin{CD} X_{Y'} @>{f_{y'}}>> Y'\\ @VV{p_{2}}V @VV{g}V\\ X @>{f}>> Y \end{CD}.
But I don't know where to use the universal property of the fiber product.