Commuting morphism of schemes

Question

If I have a diagram of morphism of schemes which commutes at level of global section, does the diagram commute?

Diagram commutes at level of topological spaces.

Answer 1

It seems this problem is equivalent to asking if $f,f':X\to Y$ are two morphisms of schemes which are equal at the levels of topological spaces and global sections, if we can conclude $f=f'$. At least any such example leads to a counterexample to your question via the diagram:

$\require{AMScd}$ \begin{CD} X @>{f} >> Y\\ @VV{\mathrm{id}}V @VV\mathrm{id}V\\ X @>{f'} >> Y \end{CD}

You can get a counterexample as follows: recall if $A$ is a ring of characteristic $p$ then the Frobenius morphism $a\mapsto a^p$ is a ring homomorphism $A\to A$ which induces the identity on the underlying topological space of $\operatorname{Spec}(A)$. By a gluing argument you can conclude that any scheme $X/\Bbb F_p$ has a "Frobenius" endomorphism $F:X\to X$ which is the identity at the level of topological spaces and has map on global sections given by $f\mapsto f^p$.

Now for the counterexample take $X=Y=\Bbb P^1_{\Bbb F_p}$, i.e. the projective line over $\Bbb F_p$. It is a standard fact that $\Gamma(\Bbb P^1_k)=k$ for any field $k$. Thus $\Gamma(X)=\Bbb F_p$, and it follows that the Frobenius endomorphism for $X$ is the identity on global sections. And it is also the identity at the level of topological spaces, but it is not the identity morphism.