Let $V_1,V_2$ be vector spaces over a same field and $W_1,W_2$ be respectively their subspaces.

Let $\mathcal V: =\{T: V_1\to V_2 \mid T(W_1)\subseteq W_2\}$.

Then there is a map $\phi: \mathcal V \to \mathcal L (V_1/W_1, V_2/W_2)$ sending $T\in \mathcal V$ to the map $\phi(T)=\tilde T: V_1/W_1 \to V_2/W_2$ defined as $\tilde T(v+W_1)=T(v)+W_2$.

My question is: is this map $\phi: \mathcal V \to \mathcal L (V_1/W_1, V_2/W_2)$ surjective ?

NOTE: For vector spaces $V,W$ by $\mathcal L(V,W)$ we mean the space of linear maps $V\to W$.

In case it helps, $\phi$ is obviously linear and I've calculated that $\ker \phi =\bigl\{T\in \mathcal V \subseteq \mathcal L(V_1,V_2) \mid T(V_1)\subseteq W_2\bigr\}$


1 Answer 1


Yes, it is. Here is a sketch of the proof:

We'll denote $p_1, p_2$ the canonical maps from $V_1$ (resp. $V_2$) onto $V_1/W_1$ (resp. $V_2/W_2$).

We can write $V_1=W_1\oplus W'_1$, where the subspace $W'_1$ is isomorphic to the quotient $V_1/W_1$.

Similarly, $V_2=W_2\oplus W'_2$, where $W'_2\overset{\varphi}{\simeq} V_2/W_2$.

Let $\tau:V_1/W_1\longrightarrow V_2/W_2$ be a linear map, and consider the section of $p_2$: $$V_2/W_2\xrightarrow{\enspace\varphi^{-1}\enspace} W'_2\xrightarrow{\enspace i\enspace\;}W_2\oplus W'_2=V_2$$ where $i$ is the canonical injection from $W'2$ into $W_2\oplus W'_2$. Set $s=i\circ\varphi^{-1}$. The linear map $$T=s\circ \tau\circ p_1:V_1\longrightarrow V_2$$ vanishes on $W_1$ and satisfies $\: \overset{\sim}{T}=\tau $.

  • $\begingroup$ I guess wherever you have tensor product, you really want direct sum ... $\endgroup$
    – user102248
    Jan 13, 2020 at 2:48
  • $\begingroup$ @user102248: You guessed right. Probably I had to tackle a problem with tensors before typing… Thank you for pointing it! $\endgroup$
    – Bernard
    Jan 13, 2020 at 10:12

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