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  "body": "\n## Proof\n\nWe build the map $V / W \\to V / U$ with the universal property and apply the first isomorphism theorem.\n\n1. 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 $x - y \\in X$.\n2. 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 $x + X = y + X$.\n3. In particular, the zero vector $0 + X$ of $V / X$ equals $x + X$ exactly when $x \\in X$.\n4. Let $U / W = \\{u + W : u \\in U\\}$. It is a [[subspace|00YC1D652K]] of the [[quotient space|0BR3FY5Q8M]] $V / W$. It contains $0 + W$, and $(u + W) + a(u' + W) = (u + au') + W$ with $u + au' \\in U$.\n5. Let $\\pi_U \\colon V \\to V / U$ be the [[quotient map|0N4F5FMQZW]], $\\pi_U(v) = v + U$. It is a [[linear map|0NRYHJPQG2]] by the operations of $V / U$.\n6. By step 3, the [[kernel|0S23E6YE83]] of $\\pi_U$ is $U$, which contains $W$.\n7. By the [[universal property of the quotient|0ZERJX6PH6]], there is a linear map $\\varphi \\colon V / W \\to V / U$ with $\\varphi(v + W) = v + U$ for all $v \\in V$.\n8. $\\varphi$ is surjective, since each $v + U$ equals $\\varphi(v + W)$. So the [[image|0WSFKKT6FF]] of $\\varphi$ is $V / U$.\n9. The kernel of $\\varphi$ is $U / W$. If $u \\in U$, then $\\varphi(u + W) = u + U = 0 + U$ by step 3.\n10. Conversely, let $\\varphi(v + W) = 0 + U$. Then $v \\in U$ by step 3, so $v + W \\in U / W$.\n11. By the [[first isomorphism theorem|0MTBGVZ4XJ]], the map $(V / W)/\\ker \\varphi \\to \\im \\varphi$, $(v + W) + \\ker \\varphi \\mapsto \\varphi(v + W)$, is well defined and a linear isomorphism.\n12. 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.\n\n## Idea\n\nSince $W \\subseteq U$, cosets of $W$ are finer than cosets of $U$, so each coset $v + W$ lies in a single coset $v + U$. This gives a surjective map $V / W \\to V / U$. It sends $v + W$ to zero exactly when $v \\in U$, so its kernel is $U / W$.\n",
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