= Solution
{c}
Apply the surjectivity of $f(Y)$ to the <identity morphism> $1_Y$. It gives a <morphism of schemes> $g:Y\to X$ satisfying $f\circ g=1_Y$. To obtain the other identity, compare $g\circ f$ and $1_X$ in $X(X)$. Their images under $f(X)$ agree, since
$$
f\circ(g\circ f)=(f\circ g)\circ f=f=f\circ1_X.
$$
The injectivity of $f(X)$ therefore gives $g\circ f=1_X$. Thus
$$
\boxed{f\text{ is an isomorphism of schemes, with inverse }g.}
$$
Only surjectivity for the test <scheme> $Y$ and injectivity for the test <scheme> $X$ were needed. This is the isomorphism-detection argument behind the <Yoneda lemma> for the <functor of points>.
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