Must every right-inverse of a linear transformation be a linear transformation? Let T be a linear transformation $\mathbb{R}^2 \rightarrow \mathbb{R}^3$. Let S be the right-inverse of T. Does S have to be linear transformation?
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By definition, if $S$ is a right-inverse of $T,$ then $S:\Bbb R^3\to\Bbb R^2$ has the property that for all $\vec v\in\Bbb R^2,$ we have $(T\circ S)(\vec v)=\vec v.$ However, by Rank-Nullity, there is a vector $\vec u\in\Bbb R^3$ with $\vec u\neq\vec 0_3,$ such that $S(\vec u)=\vec 0_2,$ so that since $T$ is a linear transformation, we would have $\vec u=(T\circ S)(\vec u)=T\bigl(S(\vec u)\bigr)=T(\vec 0_2)=\vec 0_3\neq\vec u.$ Thus, $T$ has no right-inverse.
On the other hand, $T$ will have left-inverses (infinitely-many of them, in fact) so long as its null space contains only $\vec{0}_2,$ and exactly one of the left-inverses (the one sending all elements outside the range of $T$ to $\vec{0}_2$) is a linear transformation. 
