Note that the order of and can be interchanged.
leoger: Doesn't at here secure life and avoid the ko?
unkx80: Self atari, already answered below.
mookie?:If white plays at , black can connect at , then if white plays at a b or c, black can play at either a or b. Black only needs to make one eye because he connected to the eye of the other black group, and he will secure that eye by playing at either a or b. I am a beginner, am i missing something?
Dieter: Good idea! But there is a catch: try both options again and count liberties.
mookie?: Oh, I see. If white plays at a or b and black takes the other, then white simply plays at d and captures that whole group. Thanks!
Again avoiding the ko yields only death for Black.
JoelR: This is one of a class of problems where it looks like the defender has a sure thing because of a set of miai plays, but the task is to find the one combination that makes the pairings false. In this case, Black wants an eye at , and can get it if he holds either a or b, and either c or d. But wait, if White takes both b and d.... I don't see this line of reasoning presented as often as it occurs to me. A good example where it is on the page is six die but eight live.
The throw-in here doesn't work - the marked point will become a real eye for the corner group, so the two disconnected black groups will have one real eye each and a shared eye - and thus live.