Re: [其他] onto 與合成函數

[OppOops (Oops)]

※ 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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→ OppOops     : 都用反證法..
推 dog2005xx   : 為何存在y s.t. g(y)不等於h(y)
→ OppOops     : y是用第一行敘述的, 所有f(x)都沒辦法map到的y
→ OppOops     : 至於g(y), h(y)是隨便取的, 條件有說是任意h,g
→ OppOops     : 所以我可造出滿足g。f = h。f的敘述後, 其他部份我
→ OppOops     : 可以任意調整
推 dog2005xx   : 謝謝