Completeness of Constraint Typing (type inference) question regarding $\sigma'$
Assuming that this is from Chapter 22 of Pierce's book, what's going on is that Pierce had to strengthen the statement of the theorem slightly to deal with all the fresh constraint variables tracked in $\mathcal{X}$. 

The naive statement of completeness  would be is that 


  If $\Gamma \vdash t:S|_{\mathcal{X}}C$ and $(\sigma, T)$ is a solution for $(\Gamma, t)$, then $(\sigma, T)$ is a solution for $(\Gamma, t, S, C)$. 


Basically, this says that if $(\sigma, T)$ is a solution to the type inference problem, then it will be a solution of the generated constraints $C$. However, the inference rules in Figure 22-1 sometimes will create some new contraint variables (for example, rule CT-App), all of which are recorded in $\mathcal{X}$.  

If the domain of $\sigma$ overlapped with the new variables constraint generation introduced, then you're in trouble, because now we have the possibility of inconsistencies between $\sigma$ and the constraints in $C$. So, instead Pierce proves that:


  If $\Gamma \vdash t:S|_{\mathcal{X}}C$ and $(\sigma, T)$ is a solution for $(\Gamma, t)$ and $\mathrm{dom}(\sigma) \cap \mathcal{X} = \emptyset$, then there exists a $\sigma'$ such that $(\sigma', T)$ is a solution for $(\Gamma, t, S, C)$ and $\sigma'/\mathcal{X} = \sigma$.


So he says that if $(\sigma, T)$ is a solution for $(\Gamma, t)$ that doesn't overlap with $\mathcal{X}$, then there is a solution $(\sigma', T)$ for $(\Gamma, t, S, C)$ such that $\sigma'$ is an extension of $\sigma$ to cover the variables in $\mathcal{X}$.