Suppose T \in \mathcal{L}(V,W). Define a \widetilde{T}: V / (null\ T) \to W such that:

so \widetilde{T} is the map that recovers the mapped result from an affine subset from the null space of the map. \widetilde{T} is well defined Same problem as that with operations on quotient space. We need to make sure that \widetilde{T} behave the same way on distinct but equivalent representations of the same affine subset. Suppose u,v \in V such that u+null\ T = v+null\ T. Because two affine subsets parallel to U are either equal or disjoint, we have that u-v \in null\ T. This means that Tu-Tv = 0 \implies Tu= Tv. So applying \widetilde{T} on equivalent representations of the same affine subset would yield the same result, as desired. \blacksquare properties of \widetilde{T} it is a linear map TBD proof. Basically just like do it inheriting operations from the operations on quotient space. it is injective We desire here that null\ \widetilde{T} = \{0\} which will tell us that \widetilde{T} is injective. Suppose some v + null\ T is in the null space of \widetilde{T}. So, we have that:

So, we have that v \in null\ T. Now, this means that v-0 \in null\ T. Because two affine subsets parallel to U are either equal or disjoint, v + null\ T = 0 + null\ T WLOG \forall v+null\ T \in null\ \widetilde{T}. This means that null\ \widetilde{T}=\{0\}, as desired. its range is equal to the map’s range \begin{equation} range\ \widetilde{T} = range\ T \end{equation} by definition of everything. V / null\ T is isomorphic to range\ T ….is this the point of this whole thing? Shown by the two sub-results above, and that injectivity and surjectivity implies invertability.