Proposition 2

Let X, Y, Z be topological spaces and let $f: X\rightarrow Y$ and $g: Y\rightarrow Z$ be continuous mappings.

a) If $f$ and $g$ are separated, then $g\circ f$ is separated.

b) If $g\circ f$ is separated, then $f$ is separated.

c) Suppose moreover that the mapping $f$ is proper and surjective. Then, if $g\circ f$ is separated, $g$ is separated.

Consider, in $X\times X$, the subspaces $\Delta_X$ (diagonal), $X\times_YX$ and $X\times_ZX$. These are respectively the sets of points $u$ of $X\times X$ such that pr$_1(u) =$ pr$_2(u),f\circ$pr$_1(u) =f\circ$pr$_2(u),g\circ f\circ$pr$_1(u) =g\circ f\circ$pr$_2(u)$. If $g\circ f$ is separated, $\Delta_X$ is closed in $X\times_ZX$, hence also in $X\times_YX$, whence b).

By prop. 1, (ii), applied to $W = X\times_ZX,g_1=f\circ$ pr$_1$ and $g_2=f\circ$ pr$_2,X\times_YX$ is closed in $X\times_ZX$ if $g$ is separated. If moreover $f$ is separated, $\Delta_X$ is closed in $X\times_YX$ (prop. 1, (i)), hence in $X\times_ZX$, whence a).

Finally, let us prove c). The mapping $(f, f): X\times X\rightarrow Y\times Y$ is proper (TG, I, p. 73, prop. 4). The subspace $X\times_ZX$ is the inverse image of $Y\times_ZY$ by $(f, f)$ ; by TG, I, p. 72, prop. 3, the mapping $u: X\times_Z$ $X\rightarrow Y\times_ZY$ deduced from $(f, f)$ is proper. Since $g\circ f$ is separated, the diagonal $\Delta_X$ is closed in $X\times_ZX$. Consequently, $u(\Delta_X)$ is closed in $Y\times_ZY$. Since $f$ is surjective, $u(\Delta_X)$ is the diagonal of $Y\times_ZY$. This shows that $g$ is separated.