[월:] 2012년 04월

22, March. Check that $ f^{-1}:B→ A$ ( $ a=f^{-1}(b)$ ⇔ $ b=f(a)$ ) is well-defined. ”Lemma 2.1” proof. ”Exercise #5” (p.21), and consider why this should be 1-1, onto. Check these. ① $ f:A→ B$ : 1-1 ⇔ \(\forall g,h:Z\to B\) ( $ fˆ g = fˆ h$ ⇒ $ g=h$ ) and ② … 더 읽기