提供一個清爽的直接證明
(<=)
令 T: Y -> {t, f} 為取值 t 的常數函數, c : Y -> {t, f} 為 f[X]
的特徵函數，也就是
c(y) = t there is x with f(x) = y
f otherwise
則
T * f (x) = t = c * f(x) {根據 T 跟 c 的定義}.
=> T = c {根據前提假設}
=> f[X] = Y {根據 c 的定義}
= f is onto.
(=>)
g(y) = g*f(x) = h*f(x) = h(y)
（因為 f 是 onto）
所以 g = h
※ 引述《OppOops (Oops)》之銘言：
: ※ 引述《dog2005xx (pp)》之銘言：
: : *(表示合成函數)
: : 題目:
: : f:A->B.
: : 假如f是onto if and only if
: : wherever C is a set and g:B->C and
: : h:B->C 是函數 such that g*f=h*f
: : it follow that g=h
: => f is onto function
: Let h,g be distinct functions,
: where h:B->C, g:B->C, and g。f = h。f
: since h,g are distinct, there exists α∈B such that g(α) ≠ h(α).
: Since f is onto, then ∃x∈A, such that f(x) = α.
: which implies (g。f)(x) ≠ (h。f)(x)
: <= For arbitrary h,g, g。f = h。f, implies g=h
: If f is not onto, then ∃y∈B such that ∀x∈A, f(x)≠y
: Let g(y) = k1 ≠ h(y) = k2, where k1,k2∈C
: (which g。f = h。f is still hold.)
: But g≠h.
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