{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-finite-group-homomorphism-kernel-01",
  "mko_id": "mko-finite-group-homomorphism-kernel",
  "title": {
    "zh-Hant": "C₄到C₂的kernel、fibers與單射性",
    "en": "Kernel, fibers and injectivity from C4 to C2"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-finite-group-homomorphism-kernel-01",
    "source": {
      "mko_id": "mko-finite-group-homomorphism-kernel",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "2d8504a53dc49db696cced0fde35020459fee128373b3d2efc52f01168374d11",
      "english_sha256": "4c8c7e9116861cd389dc48d302f74f3de3e520a7450771096acb9a4ec434b806"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "calc"
    ],
    "goal": {
      "zh-Hant": "f:C₄→C₂，f(a)=a mod2的kernel為{0,2}且正規；fibers是其cosets，f非單射。",
      "en": "For f:C4 to C2, f(a)=a mod 2, the normal kernel is {0,2}; fibers are its cosets and f is not injective."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "G={0,1,2,3}採模4加法，Q={0,1}採模2加法；f表依序[0,1,0,1]。",
          "en": "G={0,1,2,3} under addition modulo 4, Q={0,1} under addition modulo 2; f's aligned table is [0,1,0,1]."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "G=C_4,\\quad Q=C_2",
        "after_tex": "\\begin{array}{c|cccc}+&0&1&2&3\\\\\\hline 0&0&1&2&3\\\\1&1&2&3&0\\\\2&2&3&0&1\\\\3&3&0&1&2\\\\\\end{array}\\qquad\\begin{array}{c|cc}+&0&1\\\\\\hline 0&0&1\\\\1&1&0\\\\\\end{array}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個完整群表共20格。加法結合律由整數取餘下降，identity均0；G逆元表0,3,2,1，Q逆元0,1，皆兩側驗。",
          "en": "The two full tables contain 20 entries. Integer associativity descends; both identities are 0, with full inverse lists 0,3,2,1 and 0,1, valid on both sides."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "父群資格先完成，不能只給f陣列便套kernel定理。",
            "en": "Qualify both groups before applying kernel facts to a mapping array."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "f:G\\to Q",
        "after_tex": "f(0)=0,f(1)=1,f(2)=0,f(3)=1",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四個來源各有一個合法目標值，故total map。",
          "en": "Each of the four source elements has one legal target value, making a total map."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "total map尚非homomorphism，保乘法責任留下一步。",
            "en": "A total map is not yet a homomorphism; preservation remains to be checked."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S01",
          "S02"
        ],
        "before_tex": "f((a+b)\\bmod4)=((a+b)\\bmod4)\\bmod2",
        "after_tex": "\\begin{array}{c|cccc}f(a+b)&0&1&2&3\\\\\\hline 0&0&1&0&1\\\\1&1&0&1&0\\\\2&0&1&0&1\\\\3&1&0&1&0\\\\\\end{array}\\quad=f(a)+f(b)\\ (\\bmod2)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整16格左右共用此值；因4是2倍數，先mod4不改parity，所以每對皆保群運算。",
          "en": "All 16 pairs have the displayed value on both sides; reducing modulo 4 preserves parity, so every pair respects the operation."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "不是只核generators；完整公式與16格同時覆蓋全部。",
            "en": "This checks all pairs, not just generators, using both the formula and 16-entry table."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S02",
          "S03"
        ],
        "before_tex": "K=f^{-1}(\\{0_Q\\})",
        "after_tex": "K=\\{0,2\\}",
        "relation": "implies",
        "rule": "mko_definition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整mapping表中恰0與2送到目標identity0。",
          "en": "Exactly 0 and 2 map to target identity 0 in the complete mapping table."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "kernel元素在來源G，不是目標中的值0、1。",
            "en": "Kernel elements belong to source G, not the target value set."
          }
        },
        "leaf_basis": "mko_definition"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S01",
          "S04"
        ],
        "before_tex": "K=\\{0,2\\}",
        "after_tex": "0\\in K;\\quad0+0=0,0+2=2,2+0=2,2+2=0;\\quad-0=0,-2=2\\ (\\bmod4)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "全部四積與兩逆元都在K，直接認證子群。",
          "en": "All four sums and both inverses stay in K, directly qualifying the subgroup."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "加法記號的inverse是負號mod4，不是有理倒數。",
            "en": "The inverse in additive notation is negation modulo 4, not a rational reciprocal."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S01",
          "S03",
          "S05"
        ],
        "before_tex": "g=0,1,2,3,\\quad k=0,2",
        "after_tex": "\\begin{array}{c|cc}g+k-g&0&2\\\\\\hline0&0&2\\\\1&0&2\\\\2&0&2\\\\3&0&2\\end{array}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整8格共軛都等k，因此K正規，未只因命名kernel就跳過。",
          "en": "All eight conjugation entries equal k, directly proving normality rather than skipping it because of the name kernel."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S03",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "normality使用已核homomorphism模型；任意map的identity preimage未必子群。",
            "en": "The model is a qualified homomorphism; an arbitrary map's identity preimage need not be a subgroup."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S01",
          "S04",
          "S06"
        ],
        "before_tex": "f^{-1}(0)=\\{0,2\\},\\quad f^{-1}(1)=\\{1,3\\}",
        "after_tex": "f(a)=f(b)\\iff b-a\\in K;\\quad f^{-1}(0)=0+K,\\ f^{-1}(1)=1+K",
        "relation": "equivalent",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩fiber完整覆蓋G；同parity恰當差是偶類，直接證fiber-coset等價。",
          "en": "The two fibers cover G; equal parity is exactly an even residue difference, directly proving the fiber/coset equivalence."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G1",
            "S01",
            "S04",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "加法式b−a對應乘法式a⁻¹b，必用來源同一運算。",
            "en": "The additive difference b-a corresponds to a inverse b in multiplicative notation, using the same source operation."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S02",
          "S04",
          "S07"
        ],
        "before_tex": "0\\ne2,\\quad f(0)=f(2)=0",
        "after_tex": "f\\text{ not injective},\\quad K\\ne\\{0\\}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "明示collision與nontrivial kernel相符，否證單射。",
          "en": "The explicit collision agrees with a nontrivial kernel and refutes injectivity."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02",
            "S04",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "一個collision足夠否證，不能因f命中全部Q就稱單射。",
            "en": "One collision suffices; being onto Q would not imply injectivity."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S09",
        "uses": [
          "G1",
          "S03",
          "S04",
          "S07",
          "S08"
        ],
        "before_tex": "f(a)=f(b)\\iff b-a\\in K",
        "after_tex": "K=\\{0\\}\\iff f\\text{ injective}",
        "relation": "implies",
        "rule": "mko_definition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若K只有0，等像迫a=b；若f單射，f(k)=f(0)=0迫k=0。兩方向清楚。",
          "en": "If K={0}, equal images force a=b; if f is injective, f(k)=f(0)=0 forces k=0. Both directions are explicit."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S03",
            "S04",
            "S07",
            "S08"
          ],
          "explanation": {
            "zh-Hant": "此條件等價與本例都非的結果一致，未提前聲稱商同構。",
            "en": "The equivalence agrees with both conditions failing here and adds no quotient isomorphism."
          }
        },
        "leaf_basis": "mko_definition"
      }
    ],
    "result": {
      "zh-Hant": "f:C₄→C₂，f(a)=a mod2的kernel為{0,2}且正規；fibers是其cosets，f非單射。",
      "en": "For f:C4 to C2, f(a)=a mod 2, the normal kernel is {0,2}; fibers are its cosets and f is not injective."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "任意total map的identity preimage都正規。",
          "en": "Any total map has a normal identity preimage."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "C₄→C₂的g=[0,0,1,1]有preimage{0,1}，1+1=2不在內，且不保運算。",
          "en": "The map C4 to C2 with array [0,0,1,1] has identity preimage {0,1}, not closed since 1+1=2, and is not a homomorphism."
        }
      },
      {
        "claim": {
          "zh-Hant": "K={0}。",
          "en": "K={0}."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "f(2)=0亦在kernel。",
          "en": "f(2)=0 puts 2 in the kernel too."
        }
      },
      {
        "claim": {
          "zh-Hant": "f滿射所以單射。",
          "en": "f is surjective, hence injective."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "f(0)=f(2)，來源有collision。",
          "en": "f(0)=f(2) is a source collision."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "calc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "假設精確整數模算術與有限群、同態、preimage定義；本例群運算用加法記號。",
          "en": "Assume exact modular integer arithmetic and definitions of finite groups, homomorphisms and preimages; use additive notation here."
        }
      },
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "來源群與目標群都須由完整Cayley table認證，f須是complete total map並逐對滿足f(xy)=f(x)f(y)。kernel是目標identity的完整preimage，需直接驗subgroup與對所有g,k的gkg^{-1} closure。對任意a,b，f(a)=f(b)恰當a^{-1}b∈ker f，因此nonempty fibers是kernel cosets。kernel只含來源identity恰當f injective。非同態不得套kernel-normality；本節點不提前宣稱quotient或第一同構定理。",
          "en": "Both groups require complete Cayley-table qualification. f is a total mapping and must satisfy f(x*y)=f(x)*f(y) for every pair. Its kernel is the entire target-identity preimage, with subgroup and all-g conjugation closure checked directly. f(a)=f(b) exactly when a^(-1)*b belongs to the kernel, so each nonempty fiber is a kernel coset. The kernel is the source-identity singleton exactly when f is injective. Nonhomomorphisms cannot invoke these theorems; quotient/image isomorphism is not claimed here."
        },
        "source": {
          "mko_id": "mko-finite-group-homomorphism-kernel",
          "sha256": "2d8504a53dc49db696cced0fde35020459fee128373b3d2efc52f01168374d11",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "source_pairs",
        "lhs": "4*4",
        "rhs": "16",
        "step": "S03"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "兩完整加法群表與完整total homomorphism；不提前建商或第一同構。",
      "en": "Two complete additive-group tables and a total homomorphism, without quotient or first-isomorphism construction."
    },
    "md_symbols": [
      [
        "K",
        "來源kernel子群",
        "source kernel subgroup"
      ],
      [
        "f",
        "完整total homomorphism",
        "complete total homomorphism"
      ]
    ],
    "md_error_controls": [
      {
        "title": "目標identity / Target identity",
        "zh": "kernel取目標0的完整preimage，不是來源0的單一名稱。",
        "en": "The kernel is the entire preimage of target 0, not merely the source label 0."
      },
      {
        "title": "先同態 / Homomorphism first",
        "zh": "全部16對保群運算後才使用kernel結論。",
        "en": "Check all 16 operation pairs before kernel conclusions."
      }
    ],
    "md_open_items": []
  },
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}