Let $V_1,V_2\subseteq V$ be vector (sub)spaces. Is it true that $(V_1+V_2)/V_2=V_1/V_2$? I've tried showing this but I don't know if the steps are correct.

$$ \begin{align*} (V_1+V_2)/V_2&=\{v_1+v_2+V_2:v_1\in V_1, v_2\in V_2\} \\ &=\{\{v_1+v_2+\tilde{v}_2:\tilde{v}_2\in V_2\}:v_1\in V_1,v_2\in V_2\} \\ &=\{\{v_1+\bar{v}_2:\bar{v}_2\in V_2\}:v_1\in V_1\} \\ &=\{v_1+V_2:v_1\in V_1\} \\ &=V_1/V_2 \end{align*} $$

In the 3rd line I used that $v_2+\tilde{v}_2\in V_2$.

  • $\begingroup$ Seems fine to me. You should also use the fact that $0\in V_2$ for one of the inclusions, perhaps. $\endgroup$ – user228113 Aug 4 '16 at 12:30
  • $\begingroup$ Do you mean the quotient spaces? $\endgroup$ – Bernard Aug 4 '16 at 12:40
  • $\begingroup$ @Bernard Yeah I mean those to be quotient spaces. $\endgroup$ – user359278 Aug 4 '16 at 12:48
  • $\begingroup$ @G.Sassatelli I don't quite understand your comment. $\endgroup$ – user359278 Aug 4 '16 at 12:49
  • $\begingroup$ Thinking about it, $V_1/V_2$ is not particularly well-defined if $V_2\nsubseteq V_1$: it should be $V_1/(V_1\cap V_2)$. And I think I am mistaken: as sets, those two are not the same. Namely, the union of the elements of $V_1/(V_1\cap V_2)$ is $V_1$, while the union of the elements of $(V_1+V_2)/V_2$ is $V_1+V_2$. They are canonically isomorphic as vector spaces, though. $\endgroup$ – user228113 Aug 4 '16 at 12:54

It is false (and meaningless), unless $V_2\subset V_1$.

What is true is given by the Second isomorphism theorem in group theory: \begin{align*} V_1/V_1\cap V_2 &\xrightarrow{\;\sim\enspace}(V_1+V_2)/V_2\\ v_1+V_1\cap V_2& \xrightarrow{\quad\enspace} v_1+V_2 \end{align*}


If $V_2$ is not a subspace of $V_1$, then the quotient $V_1/V_2$ is usually not defined, so I would be careful with the notation $V_1/V_2$.

Instead, let $\pi \colon V \to V\!/V_2$, $v \mapsto v + V_2$ denote the canonical projection. If $U \subseteq V$ is any subspace containing $V_2$, then we have the equality $$ U\!/V_2 = \{ u + V_2 \mid u \in U \} = \{ \pi(u) \mid u \in U \} = \pi(U), $$ so we can replace the quotient $U\!/V_2$ by the subspace $\pi(U)$. This has the advantage that the image $\pi(U)$ is a well-defined subspace of $V\!/V_2$ for every subspace $U \subseteq V$, and not only those which contain $V_2$. (We may still think of $\pi(U)$ as the quotient $U\!/V_2$, even if this quotient is not defined.)

Your calculation then show that $\pi(V_1 + V_2) = \pi(V_1)$.


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