1
$\begingroup$

Let $f,g:X\longrightarrow X$ be two continuous maps. Recall that $f$ and $g$ are homotopic if there exists a continuous map $F:X\times I\longrightarrow X$ so that $F(x,0)=f(x)$ and $F(x,1)=g(x)$.

Is there a known result (in connection to homotopy or homology groups) from which we can obtain $f\simeq g$?

$\endgroup$
  • $\begingroup$ Are you asking is there a general approach as to how to show that two maps are homotopic? $\endgroup$ – RichoKicked800goals Jun 14 '18 at 12:36
  • $\begingroup$ @RichoKicked800goals Not in general. I just want to know that whether there is a known theorem about it or not. For example, if $\pi_i (f)=\pi_i (g)$ for all $i$, then $f\simeq g$? $\endgroup$ – M.Ramana Jun 14 '18 at 12:40
  • $\begingroup$ @M.Ramana No, this is not true. Take any non-contractible space with trivial homotopy and homology groups, e.g. open long line. Then it has non-homotopic maps $X\to X$ (namely the identity and a constant map) but they obviously induce the same map on both homotopy and homology groups (being all trivial). $\endgroup$ – freakish Jun 14 '18 at 12:47
  • $\begingroup$ @freakish Thank you very much for the comment. Indeed my question is that: is there any algebraic result (in term of homotopy and homology groups like above) from which one can conclude $f\simeq g$? $\endgroup$ – M.Ramana Jun 14 '18 at 12:51
  • $\begingroup$ For example, by the Withehead Theorem, if $X$ and $Y$ be two CW-complexes and $f:X\longrightarrow Y$ a continuous map so that $\pi_i (f)$ is an isomorphism for all $i$, then $X \simeq Y$ through $f$. Is there any similar result for maps? $\endgroup$ – M.Ramana Jun 14 '18 at 12:54
1
$\begingroup$

The homotopy classification of maps $X \to Y$ is clearly related to that of finding the path components of the function space $Y^X$.

However for some practical answers related to CW-complexes and cohomology I refer you to

Ellis, G.J. Homotopy classification the J.H.C. Whitehead way. Exposition. Math. 6~(2) (1988) 97--110.

A pdf is available here.

$\endgroup$
  • $\begingroup$ Thank you very much for the reference. $\endgroup$ – M.Ramana Jun 16 '18 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.