= Solution
{c}
\b[Yes.] First, the hypothesis implies that $f(T)$ is injective for every <scheme> $T$, even when $T$ is not an <affine scheme>. Indeed, suppose $a,b:T\to X$ satisfy $f\circ a=f\circ b$. Choose an <affine open cover> $T=\bigcup_\lambda V_\lambda$. On each $V_\lambda$, injectivity of $f(V_\lambda)$ gives $a|_{V_\lambda}=b|_{V_\lambda}$. By <gluing morphisms of schemes>, $a=b$.
Now choose an <affine open cover> $Y=\bigcup_i U_i$, and denote its inclusion maps by $j_i:U_i\hookrightarrow Y$. Surjectivity of $f(U_i)$ gives <morphisms of schemes> $g_i:U_i\to X$ with $f\circ g_i=j_i$. On $U_i\cap U_j$, the two restrictions have the same composite with $f$. By the injectivity just established for arbitrary test <schemes>,
$$
g_i|_{U_i\cap U_j}=g_j|_{U_i\cap U_j}.
$$
In particular, the overlaps need not be <affine schemes>. <Gluing morphisms of schemes> supplies $g:Y\to X$ with $g|_{U_i}=g_i$, and $f\circ g=1_Y$ because this holds on every $U_i$.
Finally,
$$
f\circ(g\circ f)=f=f\circ1_X.
$$
Injectivity of $f(X)$, already proved without an affineness assumption, gives $g\circ f=1_X$. Therefore
$$
\boxed{f\text{ is an isomorphism of schemes}.}
$$
This proves the relevant instance of <affine test schemes determine scheme morphisms>: the restricted <functor of points> still retains the information needed to construct the inverse.
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