# Contractibility of Finite Topological Space

• February 26th 2009, 05:43 PM
joeyjoejoe
Contractibility of Finite Topological Space
I'm looking for a way to show that a finite topological space with some prescribed topology is a contractible space. I understand the definition well enough: I need two continuous functions f,g such that their compositions are homotopic to the identity map on each space, where one of these spaces is just a 1-element finite topological space.

So as an example, say:
x = {a,b}
Tx = {empty set, {a}, x}
y = {a}
Ty = {empty set, {a}}

f: X -> Y is continuous
g: Y -> X is continuous

What would f and g be in this case to make the homotopies work? It seems like one of these functions would have to map a in Y to both a and b in X in order to make them homotopic to the identity functions? Am I not seeing something obvious here? Thanks.
• February 27th 2009, 02:55 AM
aliceinwonderland
Quote:

Originally Posted by joeyjoejoe
I'm looking for a way to show that a finite topological space with some prescribed topology is a contractible space. I understand the definition well enough: I need two continuous functions f,g such that their compositions are homotopic to the identity map on each space, where one of these spaces is just a 1-element finite topological space.

So as an example, say:
x = {a,b}
Tx = {empty set, {a}, x}
y = {a}
Ty = {empty set, {a}}

f: X -> Y is continuous
g: Y -> X is continuous

What would f and g be in this case to make the homotopies work? It seems like one of these functions would have to map a in Y to both a and b in X in order to make them homotopic to the identity functions? Am I not seeing something obvious here? Thanks.

If I understand your question correctly, the composition maps of f and g can be constant maps. You might already know that an identity map is nulhomotophic for a contractible space. Simply put, a contractible space X is homotopic equivalent to a point such that there exists a continuous map $F:X \times [0,1] \rightarrow X$ with $F(x,0) = id_x(x)=x$ and $F(x,1) = c(x)=c$, for all $x \in X$ and a fixed $c \in X$.

For $f \circ g:Y \rightarrow Y$, it maps from Y to its single element because Y is a contractible space and an identity map is homotopic to a constant map. It is similar to $g \circ f$.
• February 27th 2009, 10:15 AM
joeyjoejoe
Yeah that's what I was thinking, I guess I was just stuck on how to define f and g explicitly.

Since f: X -> Y, it would have to map both b and a in X to a in Y.
Similarly, g: Y -> X, it would have to map a in Y to both a and b in X.

I'm just confused as how to build these continuous maps, but I have an idea:

If you define f and g parametrically, say:

f(x) = {x, 0 <= t < 1}
{a, t = 1}

g would be defined similarly. I have already proved that this f is continuous, so I think I'm set.