Proof of the second isomorphism theorem for vector spaces

Proof

We apply the first isomorphism theorem to the map $U \to (U + W)/W$ that sends $u$ to its coset.

1. The sum $U + W$ is a subspace of $V$ that contains $W$. Indeed $0 = 0 + 0$, and $(u + w) + a(u' + w') = (u + au') + (w + aw')$. Each $w \in W$ equals $0 + w \in U + W$.
2. By the intersection of subspaces, $U \cap W$ is a subspace of $U$. So both quotient spaces $U / (U \cap W)$ and $(U + W)/W$ are defined.
3. For $x, y \in U + W$, we have $x + W = y + W$ if and only if $x - y \in W$. If $x + W = y + W$, then $x = x + 0 \in y + W$, so $x - y \in W$.
4. Conversely, if $x - y = w_0 \in W$, then $x + w = y + (w_0 + w)$ and $y + w = x + (w - w_0)$ for each $w \in W$. So $x + W = y + W$.
5. Define $\varphi \colon U \to (U + W)/W$ by $\varphi(u) = u + W$. By the operations of the quotient, $\varphi(u + au') = (u + W) + a(u' + W)$, so $\varphi$ is a linear map.
6. $\varphi$ is surjective. Let $(u + w) + W$ be an element of $(U + W)/W$, with $u \in U$ and $w \in W$. Since $(u + w) - u = w \in W$, step 4 gives $(u + w) + W = u + W = \varphi(u)$.
7. The zero vector of $(U + W)/W$ is $0 + W$. By steps 3 and 4, $\varphi(u) = 0 + W$ if and only if $u \in W$.
8. So the kernel of $\varphi$ is $\{u \in U : u \in W\} = U \cap W$. By step 6, the image of $\varphi$ is $(U + W)/W$.
9. By the first isomorphism theorem, the map $U / \ker \varphi \to \im \varphi$, $u + \ker \varphi \mapsto \varphi(u)$, is well defined and a linear isomorphism.
10. By step 8, this is the map $U / (U \cap W) \to (U + W)/W$, $u + (U \cap W) \mapsto u + W$. So this map is well defined and a linear isomorphism.