Induced linear map on quotient space

Question

Let $V$ be a vector space and $W$ be a subspace of $V$. If $J:V\to V$ is a linear map such that $J(W)\subset W$ then $J$ induces a linear map $\tilde{J}:V/W\to V/W$ with $\tilde J(v+W):=J(v)+W$

Now if we are given $V_0$ is a subspace of $V$ and define $W_0:=V_0\cap W $.

If $ J_0:=J|_{V_0}$. Then is $\tilde {J_0}:V_0/W_0\to V_0/W_0$ given by $\tilde {J_0}(v_0+W_0)= J_0(v_0)+J_0(W_0)$?

Answer 1

There may be no such map $\tilde {J_0} \colon V_0/W_0 \to V_0/W_0$, because $V_0$ (and hence $W_0$) need not be invariant under $J_0$ (under $J$).

Example

Let $V = \mathbb R^3$, and let $J$ be a non-trivial rotation about the $z$-axis. If we define $W$ to be the $x$-$y$ plane, $J(W) = W$. Now let $V_0$ be the $y$-$z$ plane. Clearly, $J(V) \not\subseteq V$. Then $W_0 = V_0 \cap W$, which is the $y$-axis, is also not invariant under $J$ or $J_0 = J|_{V_0}$. Since $J_0(W_0) \not\subseteq W_0$, $\tilde{J_0}$ cannot be a map from $V_0/W_0$ to itself.

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However, since $J(W) \subseteq W$, $J_0(W_0) = J_0(V_0 \cap W) \subseteq W$. (Note that $J_0 = J|_{V_0} \colon V_0 \to V$).

Therefore, we can define $\tilde{J_0} \colon V_0/W_0 \to V/W$ as $\tilde{J_0}(v_0 + W_0) = J_0(v_0) + W$. This is well defined, since $$u_0 + W_0 = v_0 + W_0 \iff u_0 - v_0 \in W_0 \implies J_0(u_0 - v_0) \in J(W_0) \subseteq J(W) \implies J_0(u_0) + W = J_0(v_0)+ W.$$