Here's what I'm trying to prove right now:
Let $V$ be a vector space over $\mathbb{F}$. Let $M$ be a linear subspace of $V$ and $N$ be a linear subspace of $M$. Prove that the mapping $x+N \mapsto x+M$ between the quotient spaces $V/N \to V/M$ is linear with kernel $M/N$. Then, deduce that:
$$\frac{V/N}{M/N} \cong V/M$$
Proof Attempt:
We have to show that the given relation $T: V/N \to V/M$ is a linear mapping. We define:
$$\forall x \in V: T(x+N) = x+M$$
This is totally-defined. To show well-definedness, let $x+N = y+N$ where $x,y \in V$. Then, $x-y \in N$. So, $x-y \in M$. Hence:
$$x + M = y+M$$
$$\iff T(x+N) = T(y+N)$$
To prove that it is linear, we need to show additivity and homogeneity.
- Proof of Additivity
Let $u,v \in V/N$. Then, $u = x+N$ and $v = y+N$ for some $x,y \in V$. Then:
$$T(u+v) = T((x+N)+(y+N)) = T((x+y)+N) = (x+y)+M = (x+M)+(y+M) = T(u) + T(v)$$
This proves additivity.
- Proof of Homogeneity
Let $\alpha \in \mathbb{F}$ and $u \in V/N$. Then, $u = x+N$ for some $x \in V$. So:
$$T(\alpha u) = T(\alpha(x+N)) = T(\alpha x +N) = \alpha x + M = \alpha (x+M) = \alpha T(u)$$
This proves homogeneity. Hence, $T$ is a linear map. To show that the kernel of $T$ is $M/N$, we have:
$$T(x+N) = x+M = \theta_V+M$$
$$\iff x \in M$$
$$\iff x+N \in M/N$$
$$\iff \ker(T) = M/N$$
Now, we notice that $T$ is surjective. By the first isomorphism theorem, it follows that:
$$\frac{V/N}{M/N} \cong V/M$$
That proves the desired result.
Does the proof above work? If it doesn't, why? How can I fix it?
Your approach is absolutely right!
I would change the part concerning $\ker T$, writing: \begin{align} x + N \in \ker T & \iff T(x+N) = 0 \in V/M \\ & \iff x+M = 0 \in V/M \\ & \iff x \in M \\ & \iff x + N \in M/N \end{align} which means $\ker T = M/N$.