Here's what I'm trying to prove:
Let $V$ be a vector space. Let $M$ and $N$ be linear subspaces of $V$. Then, it is the case that $M/(M \cap N)$ is isomorphic to $(M+N)/N$.
Proof Attempt:
Define the relation $Q: M/(M \cap N) \to (M+N)/N$ as follows:
$$\forall x \in M: Q(x+M \cap N) = x+N$$
This is totally defined so I will first show that it is well-defined. Suppose that $x+M \cap N = y + M \cap N$ for $x,y \in M$. Then, $x-y \in M \cap N$. So, $x-y \in M$ and $x-y \in N$. So:
$$x+N = y+N$$
$$\iff Q(x+M) = Q(y+M)$$
So, the function is well-defined.
We need to prove that this is linear and bijective. We will prove linearity first.
- Proof of Additivity
Let $u,v \in M/(M \cap N)$. Then, $u = x + M \cap N$ and $v = y + M \cap N$ for some $x,y \in M$. So:
$$Q(u+v) = Q((x+y)+M \cap N) = (x+y) + N = (x+N) + (y+N) = Q(u) + Q(v)$$
That proves additivity.
- Proof of Homogeneity
Let $\alpha \in \mathbb{F}$ and $u \in M/(M \cap N)$. Then, $u = x+M \cap N$ for some $x \in M$. So:
$$Q(\alpha u) = Q(\alpha x + M \cap N) = \alpha x + N = \alpha (x+N) = \alpha Q(u)$$
That proves homogeneity. Hence, $Q$ is linear.
Now, we will need to prove bijectivity.
- Proof of Injectivity
Let $u,v \in M/(M \cap N)$ such that:
$$Q(u) = Q(v)$$
Since $u = x + M \cap N$ and $v = y + M \cap N$ for some $x,y \in M$, we have:
$$x+N = y +N$$
$$\implies x-y \in N$$
So, $x-y \in M \cap N$ and that implies that:
$$x + M \cap N = y + M \cap N$$
Hence, $u = v$. That proves injectivity.
- Proof of surjectivity
Let $x+N \in (M+N)/N$, where $x \in M+N$. Then:
$$\exists x' \in M: \exists n \in N: x = x'+n$$
Since $x-x' \in N$, it follows that:
$$x+N = x'+N$$
It is also the case that $Q(x'+M \cap N) = x'+N$. So, define $x'+M \cap N$ as the preimage of $x+N$ and we are done. This proves surjectivity.
Since $Q$ is linear and bijective, it follows that it is an isomorphism between $M/(M \cap N)$ and $(M+N)/N$. That proves the desired result.
Does the proof above work? If it doesn't, why? How can I fix it?
The proof looks absolutely correct. There is also another way to prove it, probably a bit shorter. You can define $\pi: M\to (M+N)/N$ by $\pi(x)=x+N$. This is clearly a linear map. It is also surjective, because for each $m\in M,n\in N$ we have $(m+n)+N=m+N=\pi(m)$, so the image of $\pi$ is all $(M+N)/N$.
Finally, we can compute the kernel of $\pi$. For $x\in M$ we have $\pi(x)=N$ if and only if $x+N=N$, if and only if $x\in N$, if and only if $x\in M\cap N$. Hence $Ker(\pi)=M\cap N$. Then it follows from the first isomorphism theorem that $M/(M\cap N)\cong (M+N)/N$.