Proof of the third isomorphism theorem for vector spaces

Proof

We build the map $V / W \to V / U$ with the universal property and apply the first isomorphism theorem.

1. For a subspace $X$ of $V$ and $x, y \in V$, we have $x + X = y + X$ if and only if $x - y \in X$. If $x + X = y + X$, then $x = x + 0 \in y + X$, so $x - y \in X$.
2. Conversely, if $x - y = z_0 \in X$, then $x + z = y + (z_0 + z)$ and $y + z = x + (z - z_0)$ for each $z \in X$. So $x + X = y + X$.
3. In particular, the zero vector $0 + X$ of $V / X$ equals $x + X$ exactly when $x \in X$.
4. Let $U / W = \{u + W : u \in U\}$. It is a subspace of the quotient space $V / W$. It contains $0 + W$, and $(u + W) + a(u' + W) = (u + au') + W$ with $u + au' \in U$.
5. Let $\pi_U \colon V \to V / U$ be the quotient map, $\pi_U(v) = v + U$. It is a linear map by the operations of $V / U$.
6. By step 3, the kernel of $\pi_U$ is $U$, which contains $W$.
7. By the universal property of the quotient, there is a linear map $\varphi \colon V / W \to V / U$ with $\varphi(v + W) = v + U$ for all $v \in V$.
8. $\varphi$ is surjective, since each $v + U$ equals $\varphi(v + W)$. So the image of $\varphi$ is $V / U$.
9. The kernel of $\varphi$ is $U / W$. If $u \in U$, then $\varphi(u + W) = u + U = 0 + U$ by step 3.
10. Conversely, let $\varphi(v + W) = 0 + U$. Then $v \in U$ by step 3, so $v + W \in U / W$.
11. By the first isomorphism theorem, the map $(V / W)/\ker \varphi \to \im \varphi$, $(v + W) + \ker \varphi \mapsto \varphi(v + W)$, is well defined and a linear isomorphism.
12. By steps 8 to 10, this is the map $(V / W)/(U / W) \to V / U$, $(v + W) + U/W \mapsto v + U$. So this map is well defined and a linear isomorphism.