TITLE: How is the smoothness of the space of deformations related to unobstructedness?
QUESTION [12 upvotes]: As a beginning differential geometer, I've been trying to learn about deformation theory.  Other than Kodaira's book, I've found virtually no references from the point of view of differential geometry.  As such, my understanding is hazy.

I am asking this question in an attempt to clarify my vague understanding, especially in regard to the last bullet point below.  Precise definitions of terms are welcome, as are references to books or introductory papers on the subject.

My vague understanding goes like this:
We are interested in studying the "deformations" of a geometric object $X$ (e.g., a complex manifold $X$).  We can associate to $X$ its "deformation exact sequence," which amounts to a kernel-cokernel exact sequence in some cohomology theory (e.g., Cech cohomology) of some map $\varphi$ which I don't understand:
$$0 \to \text{Ker}(\varphi_*) \to H^1_X \xrightarrow{\varphi_*} H^2_X \to \text{Coker}(\varphi_*) \to 0.$$
We interpret this as follows:

$\text{Ker}(\varphi_*)$ is the space of infinitesimal (or "first-order") deformations.  An object $X$ is rigid iff $X$ has no infinitesimal deformations -- i.e., $\text{Ker}(\varphi_*) = 0$.
$\text{Coker}(\varphi_*)$ is the space of infinitesimal (or "first-order") obstructions.  An object $X$ is unobstructed iff $X$ has no obstructions -- i.e, $\text{Coker}(\varphi_*) = 0$.
Let $\mathcal{M}$ be the "moduli space of local deformations of $X$."  The "formal tangent space" $T_X\mathcal{M}$ is the space of first-order deformations -- i.e., $T_X\mathcal{M} \cong \text{Ker}(\varphi_*)$.
Somehow, in some cases, the smoothness of $\mathcal{M}$ is related to whether $X$ is unobstructed.  In this case, $\dim(\mathcal{M}) = \dim(T_X\mathcal{M}) = \dim(\text{Ker}(\varphi_*))$.

More hazy thoughts: I think in the cases I'm interested in, the map $\varphi_*$ can be regarded as the differential of some map $\varphi \colon \text{Somewhere} \to \text{Somewhere Else}$, and $\mathcal{M} = \varphi^{-1}(\text{point})$, so $\mathcal{M}$ is smooth if $\varphi$ is a submersion (meaning $\varphi_*$ is surjective, so $\text{Coker}(\varphi_*) = 0$), in which case $T_X\mathcal{M} = \text{Ker}(\varphi_*)$.  How any of this works precisely, I don't know.

REPLY [5 votes]: Consider the following very basic example. The affine scheme $\text{Spec } k[x, y]/xy$ is singular at the origin. This singularity can be detected as follows: the Zariski tangent space at the origin, which we think of as the space of $k$-algebra maps $k[x, y]/xy \to k[e]/e^2$ which reduce to $x, y \mapsto 0 \bmod e$, is 2-dimensional, since we can take
$$x \mapsto ae, y \mapsto be$$
for any $a, e$. But most of these tangent vectors don't lift to 2-jets. The space of 2-jets at the origin is the space of $k$-algebra maps $k[x, y]/xy \to k[e]/e^3$ which reduce to $x, y \mapsto 0 \bmod e$, and if $x \mapsto ae$ with $a \neq 0$ then we must have $y \mapsto be^2$ for some $b$, and vice versa. So the only tangent vectors which lift to 2-jets are those with $a = 0$ or $b = 0$. (Geometrically this should make sense since this variety is the union of the $x$- and $y$-axis and these are the tangent vectors pointing along those axes.) 
This example can be used to motivate the definition of formal smoothness in algebraic geometry. This is a condition weaker than being smooth, and it's usually stated in much more generality than we need here: all we need is that formal smoothness of a space $M$ (variety, scheme, stack, whatever) implies that any tangent vector $\text{Spec } k[e]/e^2 \to M$ lifts to a 2-jet $\text{Spec } k[e]/e^3 \to M$.
Now take $M$ to be a moduli space of whatever sort of objects you're trying to deform. A point of $M$ is an object of that sort. A tangent vector to a point is a first-order deformation of that object. And a 2-jet is a second-order deformation. So if $M$ is smooth at a point, then we expect any tangent vector to lift to a 2-jet, which is precisely the statement that first-order deformations lift to second-order deformations. 
If $X$ is an object describing a point of the moduli space $M$, a further question is what any of this business has to do with, say, the tangent bundle of $X$. The short story is that the correct definition of "tangent space of $M$ at $X$" should return an object called the tangent complex of $X$ (shifted by $1$), which is a derived version of the tangent bundle of $X$, and "$M$ is smooth at $X$" means that the tangent complex is concentrated in degree $\pm 1$ (it's one of these, not either, but don't ask me to figure out which), where it is just the usual tangent bundle. Obstructions live in the tangent complex in other degrees, so if there aren't any other degrees then there aren't any other obstructions.