TITLE: Nonsingular map of quotient vector space
QUESTION [1 upvotes]: Let $T:V\to V$ be a linear map and $W,U$ subspaces of $V$(finite dimensional ) with $W \subseteq U$. Also both spaces are $T$-invariant. If the induced map $V/W \to V/W $ is nonsingular then the induced map $V/U \to V/U $ is also nonsingular. I think the proof must be trivial but I cannot see it.

REPLY [0 votes]: Let $T_W\colon V/W\to V/W$ and $T_U\colon V/U\to V/U$ be the induced maps.
Let $x\in V$; then, by assumption, there exists $y\in V$ with $T_W(y+W)=x+W$. In other words, $T(y)-x\in W$, hence also $T(y)-x\in U$. Therefore $T_U(y+U)=T(y)+U=x+U$ and $T_U$ is surjective.
You can see it “diagrammatically”: if $\pi\colon V/W\to V/U$ is the canonical map, the following diagram is commutative
$$\require{AMScd}
\begin{CD}
V/W @>T_W>> V/W \\
@V\pi VV @VV\pi V\\
V/U @>T_U>> V/U
\end{CD}
$$
Since $\pi\circ T_W$ is surjective, also $T_U$ is surjective.
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