let X, Y be finite set ; for F being Function of X,Y st card X = card Y holds
( F is onto iff F is one-to-one )
let F be Function of X,Y; ( card X = card Y implies ( F is onto iff F is one-to-one ) )
assume A1:
card X = card Y
; ( F is onto iff F is one-to-one )
thus
( F is onto implies F is one-to-one )
( F is one-to-one implies F is onto )
thus
( F is one-to-one implies F is onto )
verumproof
assume
F is
one-to-one
;
F is onto
then
dom F,
F .: (dom F) are_equipotent
by CARD_1:60;
then A12:
card (dom F) = card (F .: (dom F))
by CARD_1:21;
assume
not
F is
onto
;
contradiction
then
not
rng F = Y
by FUNCT_2:def 3;
then
not
Y c= rng F
by XBOOLE_0:def 10;
then consider y being
set such that A13:
y in Y
and A14:
not
y in rng F
by TARSKI:def 3;
A15:
card (rng F) <= card (Y \ {y})
by A14, NAT_1:44, ZFMISC_1:40;
A16:
F .: (dom F) = rng F
by RELAT_1:146;
{y} meets Y
by ZFMISC_1:54, A13;
then A17:
Y \ {y} <> Y
by XBOOLE_1:83;
Y \ {y} c< Y
by XBOOLE_0:def 8, A17;
then
card (Y \ {y}) < card Y
by CARD_2:67;
hence
contradiction
by A1, A13, A15, A12, A16, FUNCT_2:def 1;
verum
end;