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Want to form the mapping cone of a map $f: X\rightarrow Y$ in the homotopy category.

I am hoping that some one can give easy examples to show that mapping cone $Y \cup_f CX$ does not dependent on f functorially and comments on when one should be careful about this.

I read this from Nilpotence and stable homotopy theory 2. In page 6, the notation and conventions section.

Thanks!

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2 Answers 2

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Actually it depends on what kind of objects $X$ and $Y$ are, and on what kind of morphism $f$ is. Because for instance, in the framework of $\infty$-categories, you can define cobifer functorially. See the seemingly innocuous 4.3.2.15 from Lurie’s Higher Topos Theory that allows to define cofibers functorially.

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I do not understand this question. The cone $CX$ on $X$ is a functorial construction and $$C(f) = Y \cup _f CX $$ with the canonical inclusion $i:X \to CX$. Where is written the suggestion that $C(f)$ is not functorial?

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  • $\begingroup$ Probably if you go to the homotopy category it's no longer functorial. That is, if f and g are homotopic maps then C(f) and C(g) are homotopy equivalent and so isomorphic in the homotopy category but not uniquely. $\endgroup$
    – user148177
    Dec 19, 2014 at 11:50
  • $\begingroup$ @Ronnie Brown, I edit the question and hope it makes sense now. I want to got to the homotopy category. $\endgroup$
    – user48537
    Dec 19, 2014 at 15:52
  • $\begingroup$ @user148177, the problem is that the choice of the mapping cone from isomorphic objects is not unique? Is there any way that I can identify all isomorphic objects as one, will this solve the problem? Thanks. $\endgroup$
    – user48537
    Dec 19, 2014 at 15:58
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    $\begingroup$ In a category, the only way you can consider two things to be really on the nose "the same" in a functorial sense is if they are isomorphic up to unique homotopy. The reason is this: let's say I have a functor F and I want F(X) to take value in that equivalence class of objects. The problem is which object in that equivalence class to choose; if they are isomorphic up to unique iso it doesn't matter, because I can make all my choices for F(X) for various X compatible since there is only one isomorphism between each object in the equiv class. But if not, then there's no way to do this. $\endgroup$
    – user148177
    Dec 19, 2014 at 21:48
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    $\begingroup$ Just to repeat Ronnie Brown's comment, the mapping cone only fails to be functorial if you take homotopy. The "mapping cone" construction comes up in contexts other than topology though. For example, it's an important construction in understanding derived categories. In this case the derived category is really the homotopy category of some infinity category (or dg category), which is why the mapping cone is NOT functorial in the derived category, but is functorial in the infinity-categorical version or dg-enriched version that the derived category comes from. $\endgroup$
    – user148177
    Dec 19, 2014 at 22:06

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