Let $\mathscr{P}$ be a property of morphism of schemes such that:
(a) closed immersion is $\mathscr{P}$;
(b) composition of $\mathscr{P}$-morphism is $\mathscr{P}$;
(e) if $f:X\to Y$, $g:Y\to Z$ are two morphisms with $g\circ f$ is $\mathscr{P}$ and $g$ is separated, then $f$ is $\mathscr{P}$.
I'm required to show that $$f:X\to Y \text{ is }\mathscr{P}\Rightarrow f_{\text{red}}:X_{\text{red}}\to Y_{\text{red}} \text{ is } \mathscr{P},$$ where $X_{\text{red}}$, $Y_{\text{red}}$ are the corresponding reduced schemes of $X$ and $Y$.
I saw another proof on some solutions available online, but I have a much shorter one. I wonder if it's legitimate:
There is natural map $i_X:X_\text{red}\to X$, where $i^\#_X$ should be the canonical surjection $\mathcal{O}_X(U)\to \mathcal{O}_X(U)_\text{red}$ on each open subset of $X$. Combining the facts that
- $i^\#_X$ is surjective;
- $i_X$ induce homeomorphism as topological space (Hartshorne Ex II.2.3),
we conclude that $i_X$ and $i_Y$ are closed immersion. Now, align a commutative diagram as follows:
Now, $i_Y\circ f_\text{red}=f\circ i_X$ is $\mathscr{P}$ from (a) and (b). $i_Y$ is $\mathscr{P}$ since closed immersion is separated. Hence $f_\text{red}$ is $\mathscr{P}$.
If I am correct that $i_X$ is a closed immersion, then I am pretty sure my argument will be correct. It will be great if someone can point out any mistakes I have made.
Thanks in advance for answering.