00:01
This problem, it is given that f is a continuous function from space x to space y.
00:06
Therefore, g of f is equals to x, comma, y, such that y equals to f of x.
00:16
Now, in the first problem, let us define h such that x tends to g of f by h of x equals to x, comma, f of x.
00:29
Now since we know that f tends, f such that x tends to y is continuous and the identity map, that is i such that x tends to x is also continuous.
00:44
We conclude that h such that x tends to g of f is also continuous.
00:53
This is also continuous.
00:54
Now h inverse such that g of f tends to x.
01:00
Is given by h inverse of x comma f of x is equals to x.
01:10
Therefore, this can be obtained as the restriction of the projection x cross y tends to capital x to the subspace g of f of x cross y.
01:27
Now, these projections are continuous and so the map h inverse is also continuous.
01:35
Here, let x1, x2 belongs to capital x, such that h of x1 is equal to h of x2...