If $T:V\to W$ is a surjective linear transformation, how do you construct a linear right inverse?

Question

Let $F$ be a field and let $V$ and $W$ be $F$-vector spaces. Suppose $T:V\to W$ is a surjective linear function. Is it always true that there exists a linear function $S:W\to V$ such that $T\circ S=\text{id}_W$?

I know that $T$ has a right inverse and that it must be injective, but it doesn't necessarily have to be linear. For example, consider $T:\mathbb{R}^2\to\mathbb{R}^1$ where $$T(\vec{v})=\begin{bmatrix}1 & 1\end{bmatrix}\vec{v}$$ and $S:\mathbb{R}^1\to\mathbb{R}^2$ where $$S(\vec{w})=\begin{bmatrix}1\\0\end{bmatrix}\vec{w}$$ for all $\vec{w}\neq \vec{0}\in\mathbb{R}^1$ but $S(\vec{0})=\begin{bmatrix}1\\-1\end{bmatrix}$.

Then $T$ is linear and surjective and $(T\circ S)(\vec{w})=\vec{w}$ for all $\vec{w}\in\mathbb{R}^1$, but $S$ is not linear, since $S(\vec{0})\neq \begin{bmatrix}0\\0\end{bmatrix}$.

I have been struggling to prove that $T$ has a linear right inverse if $T$ is surjective in the general case and would like some guidance. Maybe this statement isn't true in general, or its only true in the finitely generated setting. Thanks.

Answer 1

Let $\{w_1,\ldots,w_m\}$ be a basis of $W$. For each $j\in\{1,2,\ldots,m\}$, take $v_j\in V$ such that $T(v_j)=w_j$. Since $\{w_1,\ldots,w_m\}$ is linearly independent, then so is $\{v_1,\ldots,v_m\}$. If $S\colon W\longrightarrow V$ is the linear map such that $(\forall j\in\{1,2,\ldots,m\}):S(w_j)=v_j$, then$$(\forall j\in\{1,2,\ldots,m\}):(T\circ S)(w_j)=T(v_j)=w_j,$$and therefore $T\circ S=\operatorname{Id}_W$.

Answer 2

1. Select a basis $w_1,\ldots w_k$ of $T(V)$.
2. Select $v_1,\ldots v_k$ such that $T(v_i)=w_i$.
3. Define $S(w_i)=v_i$ for all the $i$’s $1,\ldots, n$. There is only one unique $S$ defined that way.
4. Verify $S$ does what you want!