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  "body": "\n## Proof\n\nWe check in turn that $\\bar T$ is well defined, linear, injective and surjective.\n\n1. Let $K = \\ker T$, the [[kernel|0S23E6YE83]] of the [[linear map|0NRYHJPQG2]] $T$. It is a [[subspace|00YC1D652K]] of $V$: $T(0) = 0$, and $T(u) = T(v) = 0$ gives $T(u + av) = 0$ for $a \\in F$.\n2. The [[image|0WSFKKT6FF]] $\\im T$ is a subspace of $W$: it contains $T(0) = 0$, and $T(u) + a T(v) = T(u + av)$ lies in it.\n3. For $x, y \\in V$, we have $x + K = y + K$ if and only if $x - y \\in K$. If $x + K = y + K$, then $x = x + 0 \\in y + K$, so $x - y \\in K$.\n4. Conversely, if $x - y = k_0 \\in K$, then $x + k = y + (k_0 + k)$ and $y + k = x + (k - k_0)$ for each $k \\in K$. So $x + K = y + K$.\n5. $\\bar T$ is well defined. If $v + K = v' + K$, then $v - v' \\in K$ by step 3, so $T(v) - T(v') = T(v - v') = 0$. Thus $T(v) = T(v')$.\n6. Each value $T(v)$ lies in $\\im T$, so $\\bar T$ is a map from the [[quotient space|0BR3FY5Q8M]] $V / K$ to $\\im T$.\n7. $\\bar T$ is linear. By the operations of $V / K$, $\\bar T((u + K) + (v + K)) = \\bar T((u + v) + K) = T(u + v) = T(u) + T(v)$. The last sum is $\\bar T(u + K) + \\bar T(v + K)$.\n8. Likewise, for $a \\in F$, $\\bar T(a(v + K)) = \\bar T(av + K) = T(av) = a\\, T(v) = a\\, \\bar T(v + K)$.\n9. $\\bar T$ is surjective. Each element of $\\im T$ is $T(v)$ for some $v \\in V$, and $T(v) = \\bar T(v + K)$.\n10. The zero vector of $V / K$ is $0 + K$, since $(0 + K) + (v + K) = v + K$ for all $v \\in V$.\n11. $\\bar T$ has trivial kernel. If $\\bar T(v + K) = 0$, then $T(v) = 0$, so $v - 0 \\in K$. By step 4, $v + K = 0 + K$.\n12. By [[the kernel criterion|0BJBYCGMWE]], $\\bar T$ is injective.\n13. By steps 7 to 12, $\\bar T$ is a bijective linear map, that is, a [[linear isomorphism|0B8T10CE7N]] from $V / \\ker T$ onto $\\im T$.\n\n## Idea\n\nTwo vectors have the same image under $T$ exactly when they differ by an element of the kernel. So $T$ gives a well-defined map on cosets of the kernel, and this map can no longer merge distinct cosets. Its image is the image of $T$.\n",
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