4. In an isosceles right $\triangle ABC$, $\angle C=90^{\circ}$. A line $l \parallel AB$ is drawn through $C$, and $F$ is a point on $l$ such that $AB = AF$. Then $\angle CFB=(\quad)$.
(A) $15^{\circ}$
(B) $75^{\circ}$
(C) $105^{\circ}$
(D) None of the above results is correct
4.D.

As shown in Figure 2, when $A F$ intersects with line segment $B C$, draw $C D \perp A B$ at $D$, and draw $A E \perp l$ at $E$. Then
$$
\begin{array}{l}
A E=C D \\
=\frac{1}{2} A B=\frac{1}{2} A F .
\end{array}
$$

In the right triangle $\triangle A E F$, we have $\angle C F A=30^{\circ}$.
Therefore, $\angle F A B=\angle C F A=30^{\circ}$.
In $\triangle A B F$, since $A B=A F$, then
$$
\angle A F B=\angle A B F=\frac{180^{\circ}-\angle F A B}{2}=75^{\circ} \text {. }
$$

Thus, $\angle C F B=\angle C F A+\angle A F B=105^{\circ}$.
As shown in Figure 3, when $A F$ does not intersect with line segment $B C$, draw $C D \perp A B$ at $D$, and draw $A E \perp l$ at $E$.

Then $A E=C D=\frac{1}{2} A B=\frac{1}{2} A F$.
In the right triangle $\triangle A E F$, we have $\angle C F A=30^{\circ}$.
Also, $\angle F C A=\angle C A B=45^{\circ}$, so,
$\angle F A C=180^{\circ}-\angle C F A-\angle F C A=105^{\circ}$.
Thus, $\angle F A B=\angle F A C+\angle C A B=150^{\circ}$.
In $\triangle A B F$, since $A F=A B$, then
$$
\angle A F B=\angle A B F=\frac{180^{\circ}-\angle F A B}{2}=15^{\circ} \text {. }
$$

Therefore, $\angle C F B=\angle C F A-\angle A F B=15^{\circ}$.
In summary, $\angle C F B$ is $105^{\circ}$ or $15^{\circ}$.