[proofplan] We prove the contrapositive: if $f$ is entire and omits two distinct values $a, b \in \mathbb{C}$, then $f$ is constant. The strategy is to reduce to the case $a = 0$, $b = 1$ by an affine transformation, then construct a holomorphic lift of $f$ through the universal covering map $\pi: \mathbb{H} \to \mathbb{C} \setminus \{0, 1\}$, where $\mathbb{H}$ denotes the upper half-plane. Since $\mathbb{C}$ is simply connected, the lift exists and defines an entire function $F: \mathbb{C} \to \mathbb{H}$. Composing with a Mobius transformation $\mathbb{H} \to B(0,1)$ yields a bounded entire function, which is constant by [Liouville's Theorem](/theorems/346). Therefore $f$ is constant. [/proofplan] [step:Reduce to the case where $f$ omits $0$ and $1$] Suppose $f: \mathbb{C} \to \mathbb{C}$ is entire and $f(\mathbb{C}) \subset \mathbb{C} \setminus \{a, b\}$ for two distinct points $a, b \in \mathbb{C}$ with $a \ne b$. Define the affine map \begin{align*} L: \mathbb{C} &\to \mathbb{C} \\ w &\mapsto \frac{w - a}{b - a}. \end{align*} Then $L$ is a biholomorphism of $\mathbb{C}$ with $L(a) = 0$ and $L(b) = 1$. The composition $h := L \circ f: \mathbb{C} \to \mathbb{C}$ is entire and satisfies $h(\mathbb{C}) \subset \mathbb{C} \setminus \{0, 1\}$. If we show $h$ is constant, then $f = L^{-1} \circ h$ is also constant. It therefore suffices to prove: every entire function $h: \mathbb{C} \to \mathbb{C} \setminus \{0, 1\}$ is constant. [/step] [step:Lift $h$ through the universal covering $\pi: \mathbb{H} \to \mathbb{C} \setminus \{0, 1\}$] The twice-punctured plane $\mathbb{C} \setminus \{0, 1\}$ admits the upper half-plane $\mathbb{H} = \{z \in \mathbb{C} : \operatorname{Im}(z) > 0\}$ as its universal cover. That is, there exists a holomorphic surjection \begin{align*} \pi: \mathbb{H} \to \mathbb{C} \setminus \{0, 1\} \end{align*} that is a covering map (a local biholomorphism with the path-lifting property). This is the modular lambda function; its construction uses the theory of elliptic modular functions. Since $h: \mathbb{C} \to \mathbb{C} \setminus \{0, 1\}$ is a continuous map from the simply connected space $\mathbb{C}$ into $\mathbb{C} \setminus \{0, 1\}$, the lifting criterion for covering spaces guarantees the existence of a continuous lift $F: \mathbb{C} \to \mathbb{H}$ satisfying $\pi \circ F = h$. Fix a basepoint $z_0 \in \mathbb{C}$ and a point $w_0 \in \mathbb{H}$ with $\pi(w_0) = h(z_0)$. Since $\mathbb{C}$ is simply connected, $h_*(\pi_1(\mathbb{C}, z_0)) = \{e\}$ is contained in $\pi_*(\pi_1(\mathbb{H}, w_0))$, so the lift $F$ exists and is unique with $F(z_0) = w_0$. Since $\pi$ is a local biholomorphism and $h$ is holomorphic, the lift $F$ is holomorphic. Indeed, for each $z \in \mathbb{C}$, the map $\pi$ restricts to a biholomorphism from a neighbourhood of $F(z)$ in $\mathbb{H}$ to a neighbourhood of $h(z)$ in $\mathbb{C} \setminus \{0,1\}$. Locally, $F = \pi^{-1}_{\text{loc}} \circ h$, which is a composition of holomorphic maps. [guided] The key fact from covering space theory used here is the lifting criterion: a continuous map $h: X \to Y$ from a connected, locally path-connected space $X$ into a space $Y$ that has a covering $\pi: \widetilde{Y} \to Y$ lifts to a map $F: X \to \widetilde{Y}$ (meaning $\pi \circ F = h$) if and only if $h_*(\pi_1(X)) \subset \pi_*(\pi_1(\widetilde{Y}))$. We verify the hypotheses: - $\mathbb{C}$ is connected and locally path-connected (it is a connected open subset of $\mathbb{R}^2$). - $\mathbb{C}$ is simply connected, so $\pi_1(\mathbb{C}, z_0) = \{e\}$, and thus $h_*(\pi_1(\mathbb{C}, z_0)) = \{e\}$. - The trivial group is contained in $\pi_*(\pi_1(\mathbb{H}, w_0))$ regardless of the covering. Therefore the lift $F: \mathbb{C} \to \mathbb{H}$ exists and is unique once we fix $F(z_0) = w_0$. Why is $F$ holomorphic, not merely continuous? The covering map $\pi: \mathbb{H} \to \mathbb{C} \setminus \{0,1\}$ is a local biholomorphism: for each $w \in \mathbb{H}$, there exists an open neighbourhood $V \ni w$ such that $\pi|_V: V \to \pi(V)$ is a biholomorphism. Since $F$ is continuous, $F^{-1}(V)$ is open, and on $F^{-1}(V)$ we have $F = (\pi|_V)^{-1} \circ h$. Both $(\pi|_V)^{-1}$ and $h$ are holomorphic, so $F$ is holomorphic on $F^{-1}(V)$. Since these sets cover $\mathbb{C}$, $F$ is entire. The deep input here is the existence of the covering $\pi: \mathbb{H} \to \mathbb{C} \setminus \{0,1\}$. This is a nontrivial fact from the theory of modular functions. The map $\pi$ is (a variant of) the modular lambda function $\lambda: \mathbb{H} \to \mathbb{C} \setminus \{0,1\}$, which is invariant under the congruence subgroup $\Gamma(2) \le SL_2(\mathbb{Z})$ acting on $\mathbb{H}$ by Mobius transformations. The quotient $\Gamma(2) \backslash \mathbb{H}$ is biholomorphic to $\mathbb{C} \setminus \{0,1\}$, and the projection $\mathbb{H} \to \Gamma(2) \backslash \mathbb{H}$ is the desired covering map. [/guided] [/step] [step:Map $\mathbb{H}$ into $B(0,1)$ and apply Liouville's Theorem to conclude $f$ is constant] The Mobius transformation \begin{align*} \varphi: \mathbb{H} &\to B(0, 1) \\ w &\mapsto \frac{w - i}{w + i} \end{align*} is a biholomorphism from the upper half-plane $\mathbb{H}$ onto the open unit disc $B(0, 1)$. To verify: for $w \in \mathbb{H}$, write $w = x + iy$ with $y > 0$. Then \begin{align*} |\varphi(w)|^2 = \frac{|w - i|^2}{|w + i|^2} = \frac{x^2 + (y-1)^2}{x^2 + (y+1)^2} = \frac{x^2 + y^2 - 2y + 1}{x^2 + y^2 + 2y + 1} < 1, \end{align*} since $y > 0$ implies the numerator is strictly less than the denominator. So $\varphi(\mathbb{H}) \subset B(0,1)$. Define $G := \varphi \circ F: \mathbb{C} \to B(0, 1)$. Then $G$ is entire (as a composition of holomorphic maps) and bounded ($|G(z)| < 1$ for all $z \in \mathbb{C}$). By [Liouville's Theorem](/theorems/346), $G$ is constant. Since $\varphi$ is injective, $F = \varphi^{-1} \circ G$ is constant. Since $\pi$ is a function, $h = \pi \circ F$ is constant. Since $L$ is a biholomorphism, $f = L^{-1} \circ h$ is constant. This contradicts the hypothesis that $f$ is nonconstant. Therefore, a nonconstant entire function cannot omit two values, and $f(\mathbb{C})$ omits at most one point of $\mathbb{C}$. [guided] The final step synthesises the entire argument. We have constructed a chain of holomorphic maps: \begin{align*} \mathbb{C} \xrightarrow{F} \mathbb{H} \xrightarrow{\varphi} B(0,1). \end{align*} The composition $G = \varphi \circ F$ is entire and takes values in $B(0,1)$, so $|G(z)| < 1$ for all $z \in \mathbb{C}$. In particular, $|G(z)| \le 1$ for all $z$, so $G$ is a bounded entire function. [Liouville's Theorem](/theorems/346) states that a bounded entire function is constant. We verify its hypothesis: $G: \mathbb{C} \to \mathbb{C}$ is entire and $|G(z)| \le 1$ for all $z \in \mathbb{C}$. Therefore $G$ is constant. Unwinding the chain: $G$ constant implies $F$ constant (since $\varphi$ is injective), which implies $h = \pi \circ F$ constant, which implies $f = L^{-1} \circ h$ constant. The sharpness of the result is worth noting. The theorem says "at most one point" can be omitted — can a nonconstant entire function omit exactly one point? Yes: $e^z: \mathbb{C} \to \mathbb{C}$ is entire, nonconstant, and omits $0$. So the bound "at most one" is tight. What fails if $f$ omits only one value? Say $f(\mathbb{C}) \subset \mathbb{C} \setminus \{a\}$. The once-punctured plane $\mathbb{C} \setminus \{a\}$ is not simply connected (its fundamental group is $\mathbb{Z}$), so it does admit a nontrivial universal cover, which is again $\mathbb{C}$ (via $z \mapsto a + e^z$). But a holomorphic lift $F: \mathbb{C} \to \mathbb{C}$ with $a + e^{F(z)} = f(z)$ is entire and unrestricted in its image — there is no reason for $F$ to be bounded, so Liouville's Theorem does not apply. This is precisely why the argument requires two omitted values: $\mathbb{C} \setminus \{0,1\}$ has a hyperbolic universal cover ($\mathbb{H}$, which is conformally equivalent to the disc), whereas $\mathbb{C} \setminus \{0\}$ has a parabolic universal cover ($\mathbb{C}$, which is not conformally equivalent to the disc). [/guided] [/step]