Monomorphism cancellation property

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Discussion Overview

The discussion revolves around the monomorphism cancellation property in the context of topology and homotopy theory. Participants explore whether the condition of homotopy equivalence of compositions implies homotopy equivalence of the original maps, particularly focusing on injections and split monomorphisms.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Tom questions whether the homotopy equivalence of compositions, given an injection, implies the homotopy equivalence of the original maps.
  • One participant suggests that if the map f is continuous with a continuous inverse, then homotopy equivalence of g and h follows.
  • Another participant inquires about the implications when f is not a split monomorphism and asks for counterexamples.
  • There is a mention that any two maps to a contractible space are homotopic, which raises the possibility of a counterexample in this context.
  • One participant proposes that if f is an injection into a contractible space and g is not homotopic to h, then f composed with g and h would still be homotopic.
  • A question is raised about what constitutes a "homotopy monomorphism" and whether injectivity on homotopy groups could be a sufficient condition.
  • Another participant discusses the implications of homotopy groups and injectivity, noting that while g_* = h_* does not imply g ≈ h, they seek examples where g and h are not homotopic despite their compositions being homotopic.
  • One participant references Whitehead's theorem and its implications for homotopy equivalence, expressing uncertainty about its application in this context.
  • There is a clarification attempt regarding the conditions under which the cancellation property might hold, specifically relating to injectivity of induced maps on homotopy groups.

Areas of Agreement / Disagreement

Participants express differing views on the implications of homotopy equivalence in this context, with no consensus on whether the cancellation property holds under the discussed conditions. Multiple competing perspectives remain regarding the nature of monomorphisms and homotopy.

Contextual Notes

Limitations include the dependence on specific properties of the maps involved, such as continuity and the nature of the spaces considered (e.g., contractible spaces, CW complexes). The discussion does not resolve the conditions required for the cancellation property to hold.

tom.coyne
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Hi, I know that if I have a monomorphism f:X\rightarrow Y then for any arrows g,h:A \rightarrow X we have f \circ g = f \circ h \; \Rightarrow \; g=h

However in a topological space, if I have f to be an injection but now have f \circ g \simeq f \circ h (where \simeq denotes homotopic) then does this imply that g \simeq h?

So my question is, is this true? If not what conditions would I require to make it true?

Thanks,
Tom
 
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tom.coyne said:
Hi, I know that if I have a monomorphism f:X\rightarrow Y then for any arrows g,h:A \rightarrow X we have f \circ g = f \circ h \; \Rightarrow \; g=h

However in a topological space, if I have f to be an injection but now have f \circ g \simeq f \circ h (where \simeq denotes homotopic) then does this imply that g \simeq h?

So my question is, is this true? If not what conditions would I require to make it true?

Thanks,
Tom

if f is continuous with a continuous inverse, then you will have that g and h are homotopic to each other. ( apply f^-1 to the homotopy family )
 
So if f is a split monomorphism then this works. What about when its not a split monomorphism? Can you think of any counter examples?
 
aren't any two maps to a contractible space homotopic? does that suggest a counterexample?
 
Ok, so if f is an injection into a contractible space Y then if g \not\simeq h then we would still have f \circ g \simeq f \circ h.

Thanks!
 
so what kind of map would be a "homotopy monomorphism"? what if it were injective on homotopy groups? (and you were working with nice spaces, like simplicial complexes.)

start by proving this is necessary.
 
Last edited:
If f\circ g \simeq f \circ h \; \Rightarrow \; g \simeq h then f_*\circ g_* = f_* \circ h_* \; \Rightarrow \; g_* = h_* hence f_* is injective.

If f is a map such that f_* is monic for all homotopy groups, then if we have f \circ g \simeq f \circ h then this implies that f_* \circ g_* = f_* \circ h_* hence g_*=h_* for all homotopy groups. Now I am getting a bit stuck... I know that g_*=h_* does not imply that g \simeq h. But I can't think of an example where g \not\simeq h and f \circ g \simeq f \circ h. Am I going down the right lines here?
 
I don't follow your argument. homotopy group elements are represented by maps g,h, so injectivity seems to say that fg ≈ fh implies g ≈ h, which is the hypothesis restricted to maps g,h, of spheres.

as to the other direction, have you heard of whitehead's theorem? or is it Hurewicz' theorem? that two maps are homotopic on CW complexes iff they induce the same maps on homotopy groups. better check that, it has been over 40 years since i took homotopy theory.

...well i seem to be overstating whitehead's theorem, but maybe it can be strengthened.
 
I thought about trying to use Whitehead's Theorem. But it states that if a map between CW complexes induces isomorphism on all homotopy groups then it is a homotopy equivalence. I wasn't sure how to use this though.

Also, I am not sure I totally understood your previous answer. Were you suggesting that a way to get a cancellation property might be to try:

if f:X \rightarrow Y is a map such that f_*:\pi_k(X) \rightarrow \pi_k(Y) is injective for all k then f\circ g \simeq f \circ h \; \Rightarrow \; g \simeq h??

Or did I misunderstand your response?
Thanks for your continued help!
 
  • #10
no, the converse.
 

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