# Right homotopic maps iff chain homotopic

Assume the model structure on $Ch(R)$ (chain complexes of left modules over the ring $R$) in which fibrations are dimensionwise epimorphisms (i.e. surjections) and weak equivalences are homology isomorphisms (I don't need cofibrations so I won't describe them).

As the title suggests, I need to prove that two chain maps $f,g:B \to X$ are right homotopic $\iff$ they're chain homotopic. The "$\Longleftarrow$" side is easy, since Hovey provides a path object that does the job ($P_n:=X_n\oplus X_n\oplus X_{n+1}$ with $d(x,y,z):=(dx,dy, -dz+x-y))$.

How can I prove the "$\Longrightarrow$" side, given that the right homotopy could be realized with any path object, not necessarily the one suggested? By the way, I can show it if I assume that the homotopy is indeed obtained by using that particular path object.

Thanks in advance for any hint!!

• If $B$ is cofibrant and $X$ is fibrant, then right homotopies can be realised by any path object. This is a general fact about model categories. – Zhen Lin Jan 24 '14 at 12:24
• I knew someone would point out this fact: but unfortunately, and you should have noticed it, no assumption on $B$ is made, otherwise I would have concluded immediately – Edoardo Lanari Jan 24 '14 at 12:49
• Btw, of course, any object is fibrant – Edoardo Lanari Jan 24 '14 at 12:49
• Of course I noticed it. What I didn't notice is any suggestion of why the claim is true in general. Is it a conjecture of yours? – Zhen Lin Jan 24 '14 at 14:15
• That's certainly true, at least. It is possible that right homotopy really is the same thing as chain homotopy, but that seems unlikely to me. (For instance, in simplicial sets, all objects are cofibrant, the cofibrations are the monomorphisms, and there is a canonical cylinder $\Delta^1 \times (-)$, but left homotopy does not coincide with $\Delta^1$-homotopy.) – Zhen Lin Jan 24 '14 at 14:38

I tried using the cofibrant replacement functor, since a cofibrant domain simplify the problem (as expressed in the comments).

What I've obtained is that $f \circ q_B \sim g \circ q_B$ (chain homotopic) where $q_B$ is a trivial fibration, since now any path object works and so I can use the one suggested by the author.But unfortunately it doesn't seem to be very helpful: I would like to show that $f,g$ are chain-homotopic themselves, but if $B$ is acyclic, for example, $q_B$ could be the zero map and so not many informations will be grasped.

The implication "right homotopic implies chain homotopic" is actually false unless you assume that $B$ is a cofibrant complex. Here's an example.

Let $B$ be the complex $\Bbb Z/2$ concentrated in degree zero, and let $C$ be the following complex concentrated in (homological) degrees $0$, $1$, $2$: $$\cdots 0 \to \Bbb Z \mathop{\longrightarrow}^2 \Bbb Z \mathop{\longrightarrow}^{(1,1)} \Bbb Z/2 \times \Bbb Z/2 \to 0.$$ We have the two maps $B \to C$ given by the inclusions of the two factors of $\Bbb Z/2$, and they are not chain homotopic (there are no maps from the degree zero component of $B$ to the degree $1$ component of $C$).

I claim that these maps are actually right homotopic, however. In fact, consider the map $C \to B$ given in degree zero by the map $(n,m) \mapsto n+m$. This map is a surjection, and if you compute homology of $C$ you find that it is a quasi-isomorphism. So in fact, we have constructed maps $$B \oplus B \to C \mathop{\twoheadrightarrow}^\sim B$$ where the first map is a cofibration and the second is an acyclic fibration. This means that $C$ is a path object for $B$, and so any map $C \to Y$ is a right homotopy between the two resulting maps $B \rightrightarrows Y$. In particular, we may use the identity map $C \to C$ as a right homotopy between the two inclusions.

(As an aside, given a complex $B$, a path object carries a "universal example" of two right homotopic maps, and if they are chain homotopic there then they are chain homotopic everywhere.)

• Why is $B\oplus B \to C$ a cofibration? With this model structure they're dimensionwise split-mono with cofibrant cokernel.. Moreover, even accepting that it is a cofibration, C would be a cylinder object, not a path object, and you would obtain a left homotopy between the two inclusions. – Edoardo Lanari Jan 25 '14 at 16:54
• @Lano: You're right, this is a left homotopy. (It's definitely that, because the cokernel is a bounded complex of free modules.) I'll see if I can correct my misstep. – Tyler Lawson Jan 25 '14 at 20:24
• Ok, you've convinced me about the fact that it is indeed a cofibration but the left/right problem remains.. – Edoardo Lanari Jan 25 '14 at 20:32