Solving Homotopy Classes: Proving [I,Y] Consists of Single Element

[jmjlt88]

The problem asks to show that if the space Y is path connected, then [I,Y], the set of homotopy classes of maps I into Y where I = [0,1], consists of a single element. What I tried to do is take two arbitrary continuous maps f,g: I -> Y and show that they are homotopic. For each s ε I, f(s) and g(s) are elements of Y. Thus, by our assumption, there exists some path, call it p_s, such that p_s(0)=f(s) and p_s(1)=g(s). Define F: I X I -> Y by the equation

F(s,t)=p_s(t)​

for each s ε I.
Then,

F(s,0)=p_s(0) = f(s) and F(s,1)=p_s(1)=g(s)​

for each s.

I have since learned that this is incorrect as the correct approach would be to take an arbitrary continuous map and show that it is homotopic to a constant map. My question is regarding where my argument failed. Any help/guidance would be great! Thank you!

 

[Dick]

The problem asks to show that if the space Y is path connected, then [I,Y], the set of homotopy classes of maps I into Y where I = [0,1], consists of a single element. What I tried to do is take two arbitrary continuous maps f,g: I -> Y and show that they are homotopic. For each s ε I, f(s) and g(s) are elements of Y. Thus, by our assumption, there exists some path, call it p_s, such that p_s(0)=f(s) and p_s(1)=g(s). Define F: I X I -> Y by the equation

F(s,t)=p_s(t)​

for each s ε I.
Then,

F(s,0)=p_s(0) = f(s) and F(s,1)=p_s(1)=g(s)​

for each s.

I have since learned that this is incorrect as the correct approach would be to take an arbitrary continuous map and show that it is homotopic to a constant map. My question is regarding where my argument failed. Any help/guidance would be great! Thank you!

What's wrong with that is that I don't see how you can claim that F is continuous in the two variables s and t. You picked the $p_s$ independently for each s. How do you know they fit together to make a continuous map from IxI to Y?

 

[jmjlt88]

Thank you! :) I knew it didn't quite make sense, which is why I eventually gave up and searched for a solution.