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Use the graph morphism of schemesIt is a closed immersion: the square with ${\Delta }_g:Y\to Y{\times }_{Z}Y$ and the map $X{\times }_{Z}Y\to Y{\times }_{Z}Y$, $(x,y)\mapsto (f(x),y)$, is a fibre product of schemes square. The separatedness of $g$ makes ${\Delta }_g$ a closed immersion, and base change preserves a closed immersion.

The projection $\pi :X{\times }_{Z}Y\to Y$ is the base change of the proper morphism $gf:X\to Z$, so it is a proper morphism. The closed immersion ${\Gamma }_f$ is a finite morphism, hence a proper morphism by part (i). Since $f=\pi {\Gamma }_f$, part (ii) provesThis argument supplies finite type for $f$ as well; it need not be separately assumed.

Since $g$ is a separated morphism, the diagonal morphism ${\Delta }_g:Y\to Y{\times }_{Z}Y$ is a closed immersion. The squareis cartesian. Indeed, on test schemes its pullback condition is $(f(x),y)=(t,t)$, forcing $y=t=f(x)$, and leaving exactly the choice of $x$. The graph morphism of schemes ${\Gamma }_f$ is therefore a base change of a morphism of schemes of ${\Delta }_g$. Part (c) shows that ${\Gamma }_f$ is a closed immersion.

Now assume additionally that $h=g\circ f:X\to Z$ is a closed immersion. The projection $p_Y:X{\times }_{Z}Y\to Y$ is its base change of a morphism of schemes along $g$, so is a closed immersion by part (c). Sincepart (b) shows that $f$ is a closed immersion. This covers the further assumption in the final unheaded paragraph.