- #1
BiGyElLoWhAt
Gold Member
- 1,622
- 131
The source I'm using is:
http://inperc.com/wiki/index.php?title=Homology_classes
And they say
Symmetry: A∼B⇒B∼A . If path q connects A to B then p connects B to A ; just pick p(t)=q(1−t),∀t .
Transitivity: A∼B , B∼C⇒A∼C . If path q connects A to B and path p connects B to C then there is a path r that connects A to C ; just pick:
##r(t)= \left [ \begin{array}{c}
q(2t) & \text{for} & t∈[0,1/2],\\
p(2t−1) & \text{for} & t∈[1/2,1] \\
\end{array} \right ] ##
and the problem I'm having is with the symmetry part.
Using their function definition, p(t) = q(1-t) for all t:
Let the two points A, and B, in question be (0,0) and (1,1), respectively, connected by the path q(t) = t
Then the function p(t)=q(1-t) should give me a line from (1,1) to (0,0) (B->A), but plugging it in gives me p(t) = q(1-t) = 1-t.
Neither point is on that line. I'm assuming I'm missing something, here, but I don't know what.
Thanks!
*I should add, that it makes sense to me, conceptually, but this example just doesn't seem to work for me, and I want to make sure I have a solid foundation before continuing.
http://inperc.com/wiki/index.php?title=Homology_classes
And they say
Symmetry: A∼B⇒B∼A . If path q connects A to B then p connects B to A ; just pick p(t)=q(1−t),∀t .
Transitivity: A∼B , B∼C⇒A∼C . If path q connects A to B and path p connects B to C then there is a path r that connects A to C ; just pick:
##r(t)= \left [ \begin{array}{c}
q(2t) & \text{for} & t∈[0,1/2],\\
p(2t−1) & \text{for} & t∈[1/2,1] \\
\end{array} \right ] ##
and the problem I'm having is with the symmetry part.
Using their function definition, p(t) = q(1-t) for all t:
Let the two points A, and B, in question be (0,0) and (1,1), respectively, connected by the path q(t) = t
Then the function p(t)=q(1-t) should give me a line from (1,1) to (0,0) (B->A), but plugging it in gives me p(t) = q(1-t) = 1-t.
Neither point is on that line. I'm assuming I'm missing something, here, but I don't know what.
Thanks!
*I should add, that it makes sense to me, conceptually, but this example just doesn't seem to work for me, and I want to make sure I have a solid foundation before continuing.
Last edited by a moderator: