Is the Albanese map a submersion? Let $X$ be a smooth projective variety and
$$
T=H^0(X,\Omega_{X}^1)^*/H_1(X,\mathbb{Z})
$$
the Albanese torus. Fix a point $p\in X$, one can construct the Albanese map $\phi:X\rightarrow T$ via
$$
q\mapsto [\alpha \mapsto \int_{p}^{q}\alpha],
$$
where $\alpha$ is an element of $H^0(X,\Omega_{X}^1)$.
Suppose $\phi(X)$ is a smooth subvariety of $T$, can we prove that $\phi$ is a submersion?
 A: Let $X$ be a surface of general type with $\dim H^0(X,\Omega_X^1)=2$, such that its Albanese map is surjective (there exist many examples of such surfaces).
Since the Kodaira dimension drops from 2 to 0, the Albanese map must be ramified, whence it is not a submersion.
A: Here's how I would think of the question: consider a tangent vector $v \in T_qX$ at a point $q$. So using calculus,
$$\phi(q+v) - \phi(q) = \bigg[ \alpha \mapsto \int_p^{q+v} \alpha - \int_p^q \alpha \bigg] = \bigg[ \alpha \mapsto \int_q^{q+v} \alpha \bigg] = \bigg[ \alpha \mapsto \alpha(q) \cdot v \bigg].$$
Here $\alpha(q) \cdot v$ should be understood as the canonical perfect pairing between $(\Omega_1X)_q$ (since $\alpha(q)$ is a cotangent vector at $q$) and $T_qX$. So what this shows is that the derivative of $\phi$ at $q$ is the map $D\phi_q : v \mapsto [\alpha \mapsto \alpha(q) \cdot v]$. So the question is, does this map have a fixed rank as $q$ varies?
Well, the condition $D\phi_q(v) = 0$ says that $v$ annihilates the image of the evaluation map $\mathrm{eval}_q : H^0(\Omega_1X) \to (\Omega_1X)_q$ that sends $\alpha \mapsto \alpha(q)$. Since the dot product is a perfect pairing, the dimension of $\ker D\phi_q$ is therefore just
$$\dim \ker D\phi_q = \dim (\Omega_1X)_q - \dim \mathrm{image}(\mathrm{eval}_q)$$
Or equivalently
$$\operatorname{rank} D\phi_q = \operatorname{rank} \mathrm{eval}_q.$$
So it seems to me that $\phi$ is a submersion onto its image if and only if $\mathrm{eval}_q$ has fixed rank for all $q$. That is, thinking of a trivial vector bundle $H^0(\Omega_1X) \times X$, the kernel of the evaluation homomorphism $$\operatorname{eval} : H^0(\Omega_1X) \times X \to \Omega_1X$$ needs to be a vector bundle. Clearly this isn't always the case if there are special points $q \in X$ where more global 1-forms vanish. But it holds for example if $\Omega_1X$ is globally generated. Note that $\Omega_1X$ can't possibly be globally generated in the example stated in the other answer ($X$ a surface of general type with $\dim H^0(\Omega_1X) = 2$ -- then $\mathrm{eval}$ would have to be an isomorphism and $\Omega_1X$ would be a trivial bundle).
