Wikipedia provides two formulas for block-matrix inversion:
$$ {\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}}^{-1}={\begin{bmatrix}\mathbf {A} ^{-1}+\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}&-\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\\-(\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}&(\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\end{bmatrix}},$$
and
$${\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}}^{-1}={\begin{bmatrix}(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}&-(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1}\\-\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}&\quad \mathbf {D} ^{-1}+\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1}\end{bmatrix}}.$$
Is it true then that all of the following equalities are true?
\begin{align}\mathbf {A} ^{-1}+\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}=&\;(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\\ -\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}=&\; -(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1} \\ -(\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}=&\;-\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1} \\ (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}=&\; \quad \mathbf {D} ^{-1}+\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1}\end{align}
Assuming $A,B, C, D$ are fixed.
By uniqueness of the inverse, yes, those expression are equal provided those terms involved indeed exists.