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  "body": "\n## Proof\n\nWe define $\\bar T$ on cosets, check that this makes sense, and then prove uniqueness.\n\n1. For $x, y \\in V$, 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 $x - y \\in W$.\n2. 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 $x + W = y + W$.\n3. Existence. Define $\\bar T(v + W) = T(v)$ for $v \\in V$. This is well defined: if $v + W = v' + W$, then $v - v' \\in W$ by step 1.\n4. Since $W \\subseteq \\ker T$, the [[kernel|0S23E6YE83]], we get $T(v) - T(v') = T(v - v') = 0$. So $T(v) = T(v')$.\n5. $\\bar T$ is additive. By the operations of the [[quotient space|0BR3FY5Q8M]], $\\bar T((u + W) + (v + W)) = \\bar T((u + v) + W) = T(u + v) = T(u) + T(v)$.\n6. $\\bar T$ is homogeneous. For $a \\in F$, $\\bar T(a(v + W)) = \\bar T(av + W) = T(av) = a\\, T(v)$. So $\\bar T$ is a [[linear map|0NRYHJPQG2]].\n7. For $v \\in V$, the [[quotient map|0N4F5FMQZW]] gives $\\pi(v) = v + W$. So $(\\bar T \\circ \\pi)(v) = \\bar T(v + W) = T(v)$, and $\\bar T \\circ \\pi = T$.\n8. Uniqueness. Let $S \\colon V / W \\to U$ be a linear map with $S \\circ \\pi = T$.\n9. Every element of $V / W$ has the form $v + W = \\pi(v)$ for some $v \\in V$. For it, $S(v + W) = S(\\pi(v)) = T(v) = \\bar T(v + W)$.\n10. So $S = \\bar T$. Hence $\\bar T$ is the only linear map with $\\bar T \\circ \\pi = T$, and it is given by $\\bar T(v + W) = T(v)$.\n\n## Idea\n\nA map on cosets is defined through representatives, so it only makes sense if all representatives of a coset give the same value. They differ by elements of $W$, which $T$ sends to $0$. Uniqueness holds because the quotient map hits every coset.\n",
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