Restriction of quotient spaces.

Question

Let $V_{1,2}$ be two $\mathbb{R}$-vector spaces and $U_{1,2}\subset V_{1,2}$ two linear subspaces. If $f:V_{1}\to V_{2}$ is a linear map such that $f(U_{1})\to U_{2}$, it induces a well-defined map on the level of quotient spaces:

$$[f]:\frac{V_{1}}{U_{1}}\to\frac{V_{2}}{U_{2}}$$

It is easy to see that this map is ...

• ...injective if and only if $f^{-1}(U_{2})=U_{1}$.
• ...surjective if and only if $V_{2}=f(V_{1})+U_{2}$.

Now, I have the following conjecture, which I would like to either prove or disprove:

Let $V_{1,2},U_{1,2}$ be as above and $f:V_{1}\to V_{2}$ such that $[f]$ is a well-defined isomorphism. Then the restricted map $f\vert_{W_{1}}:W_{1}\to W_{2}$ induces an isomorphism

$$[f\vert_{W_{1}}]:\frac{W_{1}}{U_{1}\cap W_{1}}\to\frac{W_{2}}{U_{2}\cap W_{2}}$$

where $W_{1}:=f^{-1}(W_{2})=\{v\in V_{1}\mid f(v)\in W_{2}\}$ denotes the preimage.

Now, it is actually quite easy to show that the map $[f\vert_{W_{1}}]$ is well-defined and injective. However, surjectivity seems to be a bit weird. We have to show that

$$W_{2}=f(W_{1})+(U_{2}\cap W_{2})$$

By assumption, we know that $V_{2}=f(V_{1})+U_{2}$. Therefore,

$$W_{2}=W_{2}\cap V_{2}=W_{2}\cap (f(V_{1})+U_{2})$$

but

$$W_{2}\cap (f(V_{1})+U_{2})=\underbrace{(W_{2}\cap f(V_{1}))}_{f(W_{1})}+(W_{2}\cap U_{2})$$

is wrong in general. Is there a way to fix it?

EDIT: Let me remark: In the special case $U_{2}\subset W_{2}$ everything works perfectly fine, so I am interested in the more general case.

Answer 1

Writing $V$ for the space $V_2,$ and respectively $X,Y,Z$ for the subspaces $f(V_1),U_2,W_2,$ your problem reduces to:

Is $V=X+Y$ a sufficient condition to derive $Z\subset(Z\cap X)+Y?$

It is indeed true if $Y\subset Z,$ but false in general (you will easily find a counterexample).