TITLE: Prove a quotient space and a subspace are isomorphic
QUESTION [5 upvotes]: Let $V_1$ and $V_2$ be subspaces of a vector space $V$ such that $V=V_1+V_2$. Let $\alpha$ be a basic for $V_1 \cap V_2$. Extend $\alpha$ to a basic $\alpha \cup \alpha_1$ for $V_1$ and $\alpha \cup \alpha_2$ for $V_2$. (where $\alpha$, $\alpha_1$ and $\alpha_2$ are pairwise disjoint). Define $V'=span\,\alpha_2$.
I need to prove that $V/V_1$ is isomorphic to $V'$. I tried to prove that $V=V_1 \oplus V'$ but failed. Also I tried to construct an isomorphism but found out that it isn't well-defined. Any thoughts about this question?

REPLY [2 votes]: The idea is to use the fundamental theorem on homomorphisms: if you can find a linear map $f:V\to V$ whose image is $V'$ and whose kernel is $V_1$, then $V/V_1\cong V'$.
Assuming that $\alpha_1\cup\alpha\cup\alpha_2$ is a basis of $V$, such a map is easy to construct: make $f$ send every element of $\alpha_2$ to itself and every element of $\alpha_1\cup\alpha$ to $0$, and then extend linearly. The image is clearly the span of $\alpha_2$, which is $V'$, and the kernel is clearly the span of $\alpha_1\cup\alpha$, which is $V_1$.
The only possible issue is that this map may not be well defined if $\alpha_1\cup\alpha\cup\alpha_2$ isn't a basis. So we have to prove that it is.
It's clearly a generating set, since it contains bases of $V_1$ and $V_2$, which in turn generate $V$ due to $V=V_1+V_2$. So only linear independence remains to be shown.
For this, consider a linear combination of elements of $\alpha_1\cup\alpha\cup\alpha_2$. It can be written as $v_1+v+v_2$, where each of the vectors is in the span of the basis with the corresponding indices. We have to show that this linear combination can only be $0$ if the vectors themselves are $0$ (some care needs to be taken here: why does the actual, full linear combination have to be trivial if these three vectors are 0?). Now assume $v_1+v+v_2=0$ but $v_2\neq0$. Then $v_2=-v_1-v$. But then $v_2\in V_1$. But $v_2$ is a linear combination of elements of $\alpha_2$, whose span contains only elements of $V_2$ which are not already contained in $V_1\cap V_2$ (other than $0$). In other words, $\alpha_2$ does not span any nonzero elements of $V_1$, meaning that $v_2$ must be zero. Apply the same argument to $v_1$, and then you have that $0+v+0=0$, meaning that $v$, too, must be $0$.