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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 $f\simeq g$?

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  • $\begingroup$ Are you asking is there a general approach as to how to show that two maps are homotopic? $\endgroup$
    – RichoKicked800goals
    Jun 14, 2018 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, 2018 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, 2018 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, 2018 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, 2018 at 12:54

1 Answer 1

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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.

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  • $\begingroup$ Thank you very much for the reference. $\endgroup$
    – M.Ramana
    Jun 16, 2018 at 9:03

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