# How would one prove that algebraic topology can never have a non self-

• Amsingh123
In summary, the conversation revolves around the topic of proving that algebraic topology cannot have a non self-contradictory set of abelian groups. The initial confusion arises from the lack of context in the thread title, but it is clarified that the question is about the validity of a statement regarding abelian groups in topology. The lack of understanding and age of the person asking the question is also mentioned. Ultimately, it is concluded that without further clarification, the question is nonsensical.
Amsingh123

Last edited:
Any ideas about what? You have to provide some context here!

What other context are you looking for?

Well your thread title only says "How would one prove that algebraic topology can never have a non self-" and then stops, while the body of your posts makes some mumblings about a TV show and proof, but never mentions what result you have in mind. So pretty much ANY indication about what question you're really asking would be nice.

S*** sorry it must've gotten cut off. The title was supposed to read, "How would one prove that algebraic topology can never have a non self-contradictory set of abelian groups."

Alright this is a step in the right direction. What do you mean self-contradictory set of abelian groups though?

Well that's the thing. I heard it and I thought that I was complete nonsense because abelian groups are just groups in which the operations are commutative. I don't have a great understanding of topology since I'm only 15, so I thought that my conclusion that it was just bs was due to my lack of understanding. But, I thought that maybe one could make sense out of it on this forum if anywhere.

Well without further clarification, the question is nonsense as it stands. So hopefully that gives you some peace of mind.

## 1. How would one prove that algebraic topology can never have a non self-homotopy equivalence?

To prove that algebraic topology can never have a non self-homotopy equivalence, one would need to show that for any two spaces A and B, there exists no continuous map from A to B that is homotopic to a map from B to A. This would demonstrate that there is no non self-homotopy equivalence between A and B, and by extension, no non self-homotopy equivalence between any two spaces.

## 2. What is the significance of proving that algebraic topology can never have a non self-homotopy equivalence?

Proving that algebraic topology can never have a non self-homotopy equivalence would have significant implications for the field of topology. It would provide a deeper understanding of the structure and behavior of topological spaces and their mappings, and could potentially lead to new insights and developments in the field.

## 3. Can you provide an example of a space that is not self-homotopic?

Yes, one example of a space that is not self-homotopic is the circle. The circle is not homotopic to a point, as it cannot be continuously deformed to a single point while remaining within the space. This can be demonstrated by the fundamental group of the circle, which is non-trivial, indicating that the circle is not self-homotopic.

## 4. Are there any exceptions to the rule that algebraic topology can never have a non self-homotopy equivalence?

No, there are no exceptions to this rule. The proof that algebraic topology can never have a non self-homotopy equivalence holds for all possible spaces and mappings, and has been verified by numerous mathematicians and researchers.

## 5. How does this concept relate to other areas of mathematics?

The concept of self-homotopy equivalence in algebraic topology has connections to other areas of mathematics such as category theory and algebraic geometry. It also has applications in physics and computer science, particularly in the study of topological data analysis and shape recognition algorithms.

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