It is well know that the isomorphism classes of first order deformations of a nonsingular variety $X$ are in $1$ to $1$ correspondence with $H^1(X,\mathcal{T}\_X)$. 


It is also known that given any small extension $A' \to A$, the group $H^1(X, \mathcal{T}\_X)$ acts *transitively* on the isomorphism classes of liftings of a given deformation $\xi$ to $A'$, if it is nonempty. 
In particular, it is not always a free action and so we no longer have a $1$ to $1$ correspondence between isomorphism classes of liftings of $\xi$ and $H^1(X, \mathcal{T}\_X)$. 


Why does the action of $H^1(X, \mathcal{T}\_X)$ fail to be free over an arbitrary small extension? What exactly makes the case $A' \to A$ so different from the case $k[\epsilon] \to k$? 


For example, suppose I want to know the first order deformation of $X\_A=\mathbb{P}^1 \times A \to A$, where $A$ is a commutative algebra over $k$, are first order deformation then still classified by $H^1(X\_A, \mathcal{T}\_{X\_A/A})$? That is by the first cohomology of the relative tangent sheaf. 



Failure of $H^1(X, \mathcal{T}_X)$ to act freely on the isomorphism classes of liftings of a deformation
The small extension $k[\epsilon]\rightarrow k$ is special because there is an inclusion of $k\hookrightarrow k[\epsilon]$ (and thus $\text{Spec}k[\epsilon]\rightarrow \text{Spec}k$) which is a section to $k[\epsilon]\rightarrow k$, and hence $X$ has a natural deformation to a scheme over $\text{Spec}k[\epsilon]$ namely $X\times\_{\text{Spec}k}\text{Spec}k[\epsilon]$. In general, for an arbitrary small extension $f:A'\rightarrow A$ which may not split in this way (for instance $\mathbb{Z}/p^2\rightarrow \mathbb{Z}/p$) there is an obstruction to the existence of a deformation. This is a class in $\mathcal{O}(f)\in H^2(X,\mathcal{T}\_X)$ (I'm assuming that the kernel $A'\rightarrow A$ is principal). This means that there is a deformation $X'$ of $X$ wrt $\text{Spec}{A}\rightarrow \text{Spec}A'$ if and only if $\mathcal{O}(f)=0$. Certainly if $ H^2(X,\mathcal{T}\_X)=0$ then the obstruction is always zero. When $A'=A[I]$ ($:=A\oplus I$ with natural ring structure) for an $A$-module $I$; $f:A'\rightarrow A$ has a natural section $A\hookrightarrow A'$ and the obstruction in $H^2(X,\mathcal{T}\_X\otimes I)$ is zero since there is a trivial deformation of $X$, namely $X\times\_{\text{Spec}A}\text{Spec}A'$ (when $A=k$ and $I=k$, $A[I]=k[\epsilon]$).


For the extension $k[\epsilon]\rightarrow k$, the first order deformations are a torsor under $H^1(X,\mathcal{T}\_X)$. For an arbitrary small extension $f:A'\rightarrow A$ (with principal kernel) they are a pseudotorsor w.r.t $H^1(X,\mathcal{T}\_X)$. In other words, if the obstruction $\mathcal{O}(f)\neq 0$, there are no first order deformations. If $\mathcal{O}(f)=0$ then the first order deformations are a torsor. In particular, it is a free transitive action and there is a bijection.


Coming back to $X=\mathbb{P}^1\times \text{Spec}A$ there is a deformation to $\text{Spec}A'$, namely, $\mathbb{P}^1\times \text{Spec}A'$ hence it is the case that the first order deformations are in bijection with $H^1(X,\mathcal{T}\_X)$.


