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

\begin{align} \widetilde{T}(v+ null\ T) = Tv \end{align}

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:

\begin{equation} \widetilde{T}(v+null\ T) = Tv = 0 \end{equation}

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.