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

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

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.

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, h₁ and h₂. Then

h₁ = h₁e = h₁(gh₂) = (h₁g)h₂ = eh₂ = h₂ .

Hence, the element x ∈ G such that gx = xg = e is unique.

TIFR YEARWISE SOLUTION PAPER,CLICK HERE

ISI YEARWISE SOLUTION PAPER,CLICK HERE

**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, h₁ and h₂. Then

h₁ = h₁e = h₁(gh₂) = (h₁g)h₂ = eh₂ = h₂ .

Hence, the element x ∈ G such that gx = xg = e is unique.

TIFR YEARWISE SOLUTION PAPER,CLICK HERE

ISI YEARWISE SOLUTION PAPER,CLICK HERE