Monomorphism cancellation property

[tom.coyne]

Hi, I know that if I have a monomorphism [itex]f:X\rightarrow Y [/itex] then for any arrows [itex]g,h:A \rightarrow X[/itex] we have [itex]f \circ g = f \circ h \; \Rightarrow \; g=h [/itex]

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

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

Thanks,
Tom

 

[wisvuze]

Hi, I know that if I have a monomorphism [itex]f:X\rightarrow Y [/itex] then for any arrows [itex]g,h:A \rightarrow X[/itex] we have [itex]f \circ g = f \circ h \; \Rightarrow \; g=h [/itex]

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

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 )

 

[tom.coyne]

So if [itex]f[/itex] is a split monomorphism then this works. What about when its not a split monomorphism? Can you think of any counter examples?

 

[mathwonk]

aren't any two maps to a contractible space homotopic? does that suggest a counterexample?

 

[tom.coyne]

Ok, so if [itex]f[/itex] is an injection into a contractible space [itex]Y[/itex] then if [itex]g \not\simeq h[/itex] then we would still have [itex]f \circ g \simeq f \circ h[/itex].

Thanks!

 

[mathwonk]

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.

 

[tom.coyne]

If [itex]f\circ g \simeq f \circ h \; \Rightarrow \; g \simeq h[/itex] then [itex]f_*\circ g_* = f_* \circ h_* \; \Rightarrow \; g_* = h_*[/itex] hence [itex]f_*[/itex] is injective.

If [itex]f[/itex] is a map such that [itex]f_*[/itex] is monic for all homotopy groups, then if we have [itex]f \circ g \simeq f \circ h[/itex] then this implies that [itex]f_* \circ g_* = f_* \circ h_*[/itex] hence [itex]g_*=h_*[/itex] for all homotopy groups. Now I am getting a bit stuck... I know that [itex]g_*=h_*[/itex] does not imply that [itex]g \simeq h[/itex]. But I can't think of an example where [itex]g \not\simeq h[/itex] and [itex]f \circ g \simeq f \circ h[/itex]. Am I going down the right lines here?

 

[mathwonk]

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.

 

[tom.coyne]

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

Or did I misunderstand your response?
Thanks for your continued help!

 

[mathwonk]

no, the converse.