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  "body": "\n## Proof\n\nWe apply the first isomorphism theorem to the map $U \\to (U + W)/W$ that sends $u$ to its coset.\n\n1. The [[sum|0JMTY8MF54]] $U + W$ is a [[subspace|00YC1D652K]] of $V$ that contains $W$. Indeed $0 = 0 + 0$, and $(u + w) + a(u' + w') = (u + au') + (w + aw')$. Each $w \\in W$ equals $0 + w \\in U + W$.\n2. By [[the intersection of subspaces|0V33SPH7RR]], $U \\cap W$ is a subspace of $U$. So both [[quotient spaces|0BR3FY5Q8M]] $U / (U \\cap W)$ and $(U + W)/W$ are defined.\n3. 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 $x - y \\in W$.\n4. 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 $x + W = y + W$.\n5. 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|0NRYHJPQG2]].\n6. $\\varphi$ 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 $(u + w) + W = u + W = \\varphi(u)$.\n7. The zero vector of $(U + W)/W$ is $0 + W$. By steps 3 and 4, $\\varphi(u) = 0 + W$ if and only if $u \\in W$.\n8. So the [[kernel|0S23E6YE83]] of $\\varphi$ is $\\{u \\in U : u \\in W\\} = U \\cap W$. By step 6, the [[image|0WSFKKT6FF]] of $\\varphi$ is $(U + W)/W$.\n9. By the [[first isomorphism theorem|0MTBGVZ4XJ]], the map $U / \\ker \\varphi \\to \\im \\varphi$, $u + \\ker \\varphi \\mapsto \\varphi(u)$, is well defined and a linear isomorphism.\n10. 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.\n\n## Idea\n\nEvery coset in $(U + W)/W$ already has a representative in $U$, because the $W$ part can be dropped. Two vectors of $U$ give the same coset exactly when their difference lies in $U \\cap W$. The first isomorphism theorem turns these two facts into the isomorphism.\n",
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