00:01
Okay, so let's follow the hint and this and let's use these facts here.
00:08
Okay, first of all, what do we have? g to the negative, the inverse image of an open subset like this, or actually any subset of o, is what? is the subset of k? so, x, oh, actually is the subset of y.
00:31
So let me write, let me use lowercase y.
00:37
Consisting of those points, y belonging to capital y, such that g of y belongs to o.
00:50
But this is equivalent to saying that our point, lower case y, is such that f to the minus 1 of y belongs to o, which is equivalent to saying that y belongs to the inverse image of o via f.
01:16
And this thing shows this equation here.
01:22
Now, we know that this inverse function, f2, the minus 1, exists because f is 1 to 1 and onto.
01:34
Now we just need to show that g to the negative one is continuous...