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### Solved Problems from TIFR,ISI,CMI Papers On Topology

If you go through TIFR,ISI,CMI previous year papers you must see these 3 problems came rapidly. So, here is your needed solutions.

Q. Show that no two of the spaces (0, 1), (0, 1], and [0, 1] are homeomorphic.
Answer: Suppose (0, 1) and (0, 1] are homeomorphic, with the homeomorphism
given by f. If A = (0, 1) − {f  ¹(1)}, then f|A : A(0, 1) is a
homeomorphism. However, the interval (0, 1) is connected, whereas
A = (0,f  ¹(1) (f  ¹(1), 1)
is not connected, so A and (0, 1) cannot be homeomorphic. From this contradiction,
then, we conclude that (0, 1) and (0, 1] are not homeomorphic.

Similarly, suppose g : (0, 1][0, 1] is a homeomorphism. Then, if
B =(0, 1] − {g ¹(0), g ¹(1)}, g|B : B(0, 1) is a homeomorphism.
However, g ¹(0)g ¹(1), so at most one of these can be 1, meaning one must lie in
the interval (0, 1). Suppose, without loss of generality, that g ¹(0)(0, 1).
Then   B = (0, g ¹(0))(g ¹(0), 1] − {g ¹(1)}
is not connected, whereas (0, 1) is, so the two cannot be homeomorphic.
From this contradiction, then, we conclude that (0, 1] and [0, 1] are not
homeomorphic.
A similar argument easily demonstrates that (0, 1) and [0, 1] are not homeomorphic,
so we see that no two of the spaces (0, 1), (0, 1], and [0, 1] are
homeomorphic.

Q.Suppose f : (0, 1)  (0, 1) is a continuous map. Does f have a fixed
point?
Answer: No. As a counter-example, consider f(x) = x². For c(0, 1),
c − f(c) = c   c² = c(1 − c) < c
since 0 < 1 − c < 1. Hence, f(c)c for all c(0, 1), so f has no fixed
points.

Q. Show that for any g  G there is a unique element x  G such that
x · g = g · x = e. This element is called the inverse of g and is usually
denoted as g  ¹.
Answer: Let g  G. By definition of a group, we know there must exist at
least one element h  G such that
hg = gh = e.
Suppose there exist two such elements, h and h. Then
h = he = h(gh) = (hg)h = eh = h₂ .
Hence, the element x  G such that gx = xg = e is unique.