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Let $S$ be an arbitrary scheme, and let $X,S'$ be $S$-schemes.

Using e.g. EGA I (3.3.9), (3.3.14), one obtains for any $S'$-scheme $T$, viewed as an $S$-scheme via $S'\to S$, a canonical bijection of sets: $$f_X:Hom_{S'}(T,X_{S'})\cong Hom_S(T,X).$$

For any morphism $\varphi:X\to Y$ of $S$-schemes, let $\varphi^*:Hom_{S}(T,X)\to Hom_{S}(T,Y)$ be the induced map and let $\varphi_{S'}:X_{S'}\to Y_{S'}$ be the induced morphism of $S'$-schemes.

I suspect that the by $\varphi_{S'}$ induced map
$$\varphi_{S'}^*:Hom_{S'}(T,X_{S'})\to Hom_{S'}(T,Y_{S'})$$ satisfies $\varphi_{S'}^*=f_Y^{-1}\circ \varphi^*\circ f_X$. In case this is true, is there a simple way (which avoids working out all canonical maps involved) to see this?

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  • $\begingroup$ Draw a diagram. $\endgroup$
    – abz
    Commented Aug 16, 2013 at 15:36
  • $\begingroup$ If you drew a diagram, you would see that the first bijection follows from the definition of a fibred product (no need to refer to EGA I, 3.3.9, 3.3.14), that what you call $\varphi^*_{S'}$ is just composition with $\varphi_{S'}$, and everything should be clear to you. $\endgroup$
    – abz
    Commented Aug 16, 2013 at 23:21
  • $\begingroup$ Dear Anon, thank you very much for your comments. Unfortunately, it is still not clear to me. Would it be possible for you to draw the diagram you mean in the answer box? $\endgroup$
    – Guillaume Pastorini
    Commented Aug 20, 2013 at 10:25

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