Morphism schemes after base extension I have been reading Ravi Vakil notes on algebraic geometry, and one exercise asks if $phi:X\rightarrow Y$ and $\pi:X\rightarrow Y$ are morphisms of $k$-schemes and $\ell/k$ a field extension, if $\pi_\ell=\rho_\ell$, then show that $\pi=\rho$, where $\pi_\ell$ and $\rho_\ell$ mean the induced maps after base change.
I can seem to reduce it to the affine case, and since $X \times_k \ell\rightarrow X$ is surjective, we get (as Ravi Vakil suggests in the notes) that $\pi$ and $\rho$ agree on the level of sets. However, I cannot seem to complete the proof that they must be the same, as agreeing on the level of sets is not enough.
Thank you for any guidance, help, or advice.
 A: The crucial fact you need is that $X \otimes_k \ell \to X$ is an epimorphism in the sense of category theory, and for this, it suffices to know that $X \otimes_k \ell \to X$ is an effective epimorphism, such as a faithfully flat morphism. But the property of being faithfully flat is preserved by base change, so it suffices to verify that $\operatorname{Spec} \ell \to \operatorname{Spec} k$ is a faithfully flat morphism, and this is easy: it is obviously surjective on points, and any morphism with codomain $\operatorname{Spec} k$ is flat.
That said, the above is overkill. As you say, the problem is local on $X$ and $Y$, so we may assume that both $X$ and $Y$ are affine. The problem then reduces to the following statement in commutative algebra:

If $f, g : A \to B$ are $k$-algebra homomorphisms such that $f \otimes_k \ell = g \otimes_k \ell$ as $\ell$-algebra homomorphisms $A \otimes_k \ell \to B \otimes_k \ell$, then $f = g$. 

Indeed, for each $k$-algebra homomorphism we have a commutative diagram
$$\begin{array}{ccc}
A & \to & A \otimes_k \ell \\
\downarrow & & \downarrow \\
B & \to & B \otimes_k \ell
\end{array}$$
and it is easy to verify that the horizontal arrows are injective homomorphisms, so the claim follows.
