Prove a quotient space and a subspace are isomorphic 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?
 A: You should refer to the collection of "basis" vectors as a "basis" and the individual vectors in them as "basis elements" or "basis vectors".
Just follow what the question tells you. I am not assuming finite dimension.
Now you have $\alpha,\alpha_{1},\alpha_{2}$ are pairwise disjoint sets.
Let $\alpha_{1}'=\alpha\cup\alpha_{1}$ and $\alpha_{2}'=\alpha\cup\alpha_{2}$
Now any vector in $V$ can be written as $w_{1}+w_{2}$ where $v_{i}\in V_{i}$.
But $\displaystyle w_{1}=\sum_{i=1}^{n_1}c_{i}v_{i}$ for $v_{i}\in\alpha_{1}'$.
Of these say $\{v_{1},v_{2},...v_{k_1}\}\in \alpha$ and $\{v_{k_1+1},..v_{n_{1}}\}\in\alpha_{1}$
And $\displaystyle w_{2}=\sum_{i=1}^{k_2}d_{i}x_{i}+\sum_{i=k_2+1}^{n_2}d_{i}x_{i}$ such that $\{x_{1},x_{2},...x_{k_{2}}\}\in \alpha$ and $\{x_{k_{2}+1},...x_{n_2}\}\in\alpha_{2}$.
Then $$w_{1}+w_{2}=\left(\sum_{i=1}^{k_1}c_{i}v_{i}+\sum_{i=1}^{k_2}d_{i}x_{i}+\sum_{i=1}^{n_1}c_{i}v_{i}\right)+\left(\sum_{i=k_2+1}^{n_2}d_{i}x_{i}\right)$$.
So the $\left(\sum_{i=1}^{k_1}c_{i}v_{i}+\sum_{i=1}^{k_2}d_{i}x_{i}+\sum_{i=1}^{n_1}c_{i}v_{i}\right)$ is from $V_{1}$ and $\left(\sum_{i=k_2+1}^{n_2}d_{i}x_{i}\right)$ is from $V'$.
Now if $v\in V\cap V'$ . Then $v\in Span(\alpha_{2})$ and $v\in Span(\alpha\cup\alpha_{1})$ .
But as $\alpha_{2},\alpha$ and $\alpha_{1}$ are pairwise disjoint. So $\alpha_{2}\cap\left(\alpha\cup\alpha_{1}\right)=\phi$ and $\alpha_{2}\cup\left(\alpha\cup\alpha_{1}\right)$ is a basis for $V$. Thus the spans are also disjoint.
Hence you have $V=V\oplus V'$.
Now define $f:V\to V'$ by $f(v_{1}+v')=v'$ such that $v_{1}\in V_{1}$ and $v'\in V'$ This map is well defined as $V=V\oplus V'$ and it is surjective. Now it is easy to see that the kernel is $V_{1}$. And hence by Isomorphism Theorem you have $V/V_{1}\cong V'$.
If you assume finite dimension then it is much easier and you can work with the explicit basis elements. But this is more general.
A: 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 $0$.
