Is There a Flaw in the Symmetry Proof for Homology Classes?

In summary, the source provided defines homology classes and discusses two properties, symmetry and transitivity. For symmetry, if A is connected to B by path q, then B can also be connected to A by path p(t) = q(1-t). For transitivity, if A is connected to B and B is connected to C, then A can be connected to C by path r(t) = q(2t) for t in [0,1/2] and p(2t-1) for t in [1/2,1]. The example given is (0,0) and (1,1) connected by path q(t) = (t,t) and path p(t) = (1
  • #1
BiGyElLoWhAt
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The source I'm using is:
http://inperc.com/wiki/index.php?title=Homology_classes [Broken]
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.
 
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  • #2
BiGyElLoWhAt said:
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.
Since you take two points in ##\mathbb R^2##, your path should also be a function into ##\mathbb R^2##.
So you have ##q(t)=(t,t)## and then ##p(t)=(1-t,1-t)##
 
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  • #3
Oh, wait, I think I get it now.
So the function needs to be on t = [0,1] and [1,0] respectively, so when you plug in 0 and 1 in q(t)=t you get 0 and 1 out, so (0,0) and (1,1) and when you plug in [1,0] into p(t)=1-t you get 1-1 = 0 and 1-0 = 1.

Is this correct?
 
  • #4
Ok, yea, for some reason I wasn't thinking parametrically D=
 

What is symmetry?

Symmetry is a mathematical concept that refers to the property of having the same shape, size, or arrangement on both sides of a dividing line or plane. In other words, it means that an object or system can be divided into two or more identical parts.

What is transitivity?

Transitivity is a mathematical property that describes the relationship between three elements or objects. It states that if one element is related to a second element, and the second element is related to a third element, then the first element must also be related to the third element.

How are symmetry and transitivity related?

Symmetry and transitivity are related because they both describe relationships between elements or objects. Symmetry describes relationships between two identical parts, while transitivity describes relationships between three elements. Additionally, both symmetry and transitivity are important concepts in fields such as algebra, geometry, and graph theory.

What are some real-life examples of symmetry and transitivity?

Some real-life examples of symmetry include the human body, snowflakes, and flowers. These objects exhibit symmetry because they can be divided into two or more identical parts. Some examples of transitivity include family relationships (e.g. if person A is the parent of person B, and person B is the parent of person C, then person A is also the grandparent of person C) and the transitive property of equality in mathematics (e.g. if a = b and b = c, then a = c).

Why are symmetry and transitivity important in science?

Symmetry and transitivity are important in science because they help us understand the relationships between different elements or objects. They also allow us to make predictions and draw conclusions based on these relationships. In fields such as biology and chemistry, symmetry and transitivity are essential for understanding the structure and function of molecules and organisms. In physics and engineering, symmetry and transitivity are crucial for understanding the laws of nature and designing efficient systems.

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