1
$\begingroup$

I am confused about Remark 2.1.11 of Riehl's categorical homotopy theory.


Let $M,N$ be homotopical categories. (Def. 2.1.1 in book).

if $F,F':M \rightarrow N$ are homotopical functors. $\alpha:F \rightarrow F'$ is a natural transformation, then this descends to a unique transformation $\delta \alpha$ between functor $\operatorname{Ho} M \rightarrow \operatorname{Ho} N$, their homotopy categories.

Here $\delta: N \rightarrow \operatorname{Ho} N, \gamma:M \rightarrow \operatorname{Ho} M$ are the localization functors with respect to weak equivalences.


I do not know how this descending and uniqueness work - simply from universal proeprty. May someone elaborate the details?

$\endgroup$

1 Answer 1

3
$\begingroup$

Recall that Riehl defines the localisation functor $\gamma_M:M\rightarrow Ho M$ to be the identity on objects, and to take a morphism $f:x\rightarrow y$ in $M$ to the equivalence class of zig-zags $$\left[x\xrightarrow{f}y\right]:x\rightarrow y$$ in the homotopy category $Ho M$.

If $F:M\rightarrow N$ is a homotopy functor, i.e. it sends weak equivalences in $M$ to isomorphisms in $N$, then we get a functor $\overline F:Ho M\rightarrow N$ defined on objects by

$$\overline F(x)=F(x)$$

and on morphisms $\varphi:x\rightarrow y$ in $Ho M$ as follows. Here $\varphi$ is an equivalence class of zig-zags in $M$, where the only maps that point in the 'wrong' direction are weak equivalences. Choose a representative for $\varphi$, say

$$x=x_0\xrightarrow{f_0}x_1\xleftarrow{w_1}x_2\xrightarrow{f_2}\dots\xleftarrow{w_{n-1}}x_{n-1}\xrightarrow{f_n}x_n=y$$

and set

$$\overline F(\varphi)=F(f_n)F(w_{n-1})^{-1}F(f_{n-1})\dots F(w_1)^{-1}F(f_0).$$

I'll leave you to check that this is well-defined on equivalence classes. Technically you should be more careful with choosing you representatives: the first and last arrows need not point in the directions I have indicated. Of course you can use the relations to arrange that any two adjacent arrows always point in the opposite directions, but you do need to check that these relations are respected by the definition of $\overline F$.

This is the first part of your question, since (hopefully) it should be pretty clear from the definition that as functors $M\rightarrow N$ the relation

$$\overline F\circ \gamma_M=F$$

holds. Morever uniquness should be clear too, since there really is no other way we can define $\overline F$ so that it has this property.

For the second part, let $\alpha:F\Rightarrow G:M\rightarrow N$ be a natural transformation of homotopy Functors. Define $\overline F,\overline G:Ho M\rightarrow N$ as above and let $\overline \alpha:\overline F\Rightarrow\overline G$ be defined componentwise by

$$\overline\alpha_x=\alpha_x:\overline F(x)=F(x)\rightarrow G(x)=\overline G(x).$$

where $x$ is an object of $M$.

To check that this is well-defined just use the definitions of $\overline F$ and $\overline G$ are defined, and the fact $\alpha$ is a natural transformation. You have to check that two types of square commute in $N$. The first do because the are of the form $\require{AMScd}$ \begin{CD} \overline F(x_i)=F(x_i)@>F(f)=\overline F(f)>> F(x_{i+1})=\overline F(x_{i+1})\\ @VV\alpha_{x_i} V @VV \alpha_{x_{i+1}}V\\ \overline G(x_i)=G(x_i) @>\overline G(f)=G(f)>> \overline G(x_{i+1})=G(x_{i+1}). \end{CD} and the second do because they are of the form $\require{AMScd}$ \begin{CD} \overline F(x_j)=F(x_j)@>\overline F(w)=F(w)^{-1}>> F(x_{j+1})=\overline F(x_{j+1})\\ @VV\alpha_{x_j} V @VV \alpha_{x_{j+1}}V\\ \overline G(x_j)=G(x_j) @>\overline G(w)=G(w)^{-1}>> \overline G(x_{i+j})=G(x_{i+j}). \end{CD} Its not so immediately obvious that these second type of diagram commute, but recall that $w$ is a weak equivalence, and the functors $F,G$ are homotopy functors. Thus the horizontal morphisms in this diagram are isomorphisms in $N$. Once you observe this it is easy to see that the square commutes.

Now check directly that the horizontal composition of natural transformations obeys the rule $$\overline\alpha\circ\gamma_M=\alpha.$$ Again, a moments reflection gives you the uniqueness statement - there really is no other way to define $\overline\alpha$ so that it would have this property!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .