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I am studying the basics of infinity categories and am rather confused about the issues of uniqueness (my main example is the $\infty$ derived category of chain complexes of vector spaces over a field of char $0$). The theorem of Joyal on uniqueness of composition seems to me to imply that a given quasi-isomorphism has a contractible space of maps homotopic to it. But I seem to also have an example where that is not true: a complex of the zero map $0: A\to A$. If I start with the identity map then the only map homotopic to it is the identity map. Homotopies are given by any map (degree 1) $A\to A$ and there are no 2-cell homotopies. What am I confused about ?

Edited: Is it possible that I am allowed to change the complex itself ?

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    $\begingroup$ It is certainly not true that the space of maps homotopic to a given equivalence is contractible; I'm not sure what led you to that conclusion. You may be confusing it with the space of maps equipped with a homotopy to a given map, which is contractible. $\endgroup$
    – Eric Wofsey
    Commented Apr 7, 2013 at 0:23
  • $\begingroup$ Thanks, you are right - what follows from Joyal's theorem is the contractibility of the space of maps equipped with a homotopy to a given map. Still... does not it imply that for the complex above all the homotopies should be equal ? And they are not ? $\endgroup$
    – lostinfinity
    Commented Apr 7, 2013 at 0:50
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    $\begingroup$ Degree 2 maps of chain complexes only correspond to homotopies between homotopies if the endpoints of the (second) homotopies are required to be fixed in place throughout the (first) homotopies. So the lack of degree 2 maps tells you that the degree 1 maps are not homotopic as homotopies whose endpoints are fixed, but if one endpoint is allowed to move they may still be homotopic. To get the space of maps equipped with a homotopy to a given map (so one endpoint but not the other is fixed), you have to do something more subtle using a simplicial structure on the category of chain complexes. $\endgroup$
    – Eric Wofsey
    Commented Apr 7, 2013 at 1:18
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    $\begingroup$ @Eric: It looks like your comments would make a good answer to this question. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 7, 2013 at 2:30
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    $\begingroup$ The higher homotopy groups of the space of maps homotopic to $f\colon A\rightarrow B$ are the sets of homotopy classes of maps $[\Sigma^nA,B]$. They do not depend on $f$. $\endgroup$
    – Fernando Muro
    Commented Apr 7, 2013 at 7:46

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