When is commuting in $\pi_1$ equivalent to commuting up to homotopy?

It seems to me that commuting up to homotopy certainly implies commuting in $$\pi_1$$, but when is the converse also true?

I am reading something on surfaces, and it says in that case commuting in $$\pi_1$$ implies commuting up to homotopy, by some "standard facts from algebraic topology", but I am unable to find it by searching the internet. Does it have anything to do with the higher homotopy groups (which I am also not very familiar with)? How about in general in higher dimensions?

Edit: I probably didn't articulate very well... I am looking into this case. If I have the following commutative diagram. $$\begin{array}{ccc} \pi_1(M)&\longrightarrow & \pi_1(M)\\ \downarrow & & \downarrow{\alpha_*}\\ \pi_1(N)&\longrightarrow & \pi_1(N) \end{array}$$ I would like to know when I can tell that I have the following diagram $$\begin{array}{ccc} M &\longrightarrow & M \\ \downarrow & & \downarrow{\beta}\\ N &\longrightarrow & N \end{array}$$ that also commutes but up to homotopy?

Additionally, suppose as in the above diagrams the corresponding maps are $$\alpha_*$$ which is induced by some map $$\alpha$$ and $$\beta$$ induces the same homomorphism on $$\pi_1$$, when can I tell $$\alpha$$ is actually homotopic to $$\beta$$? I guess this is essentially the same as asking, when are the two maps that induce the same homomorphism on $$\pi_1$$ really homotopic to each other?

I am in the case where the above maps are surjective, but I would love to know if it is true in a more general case.

• Are $M$ and $N$ closed surfaces or otherwise aspherical? In general you absolutely cannot conclude that. – user98602 Nov 8 '18 at 2:45
• Hi Mike, yes I am looking into closed surfaces! And when asking around I heard the word aspherical but the one that I asked wasn't so sure about that... I would like to know if there are exact conditions that guarantee this and where I could probably find them (a proof). Thanks!! – chikurin Nov 8 '18 at 2:50

This is about the fact that the universal cover of a closed surface of positive genus is contractible. In particular, all of its "higher homotopy groups" are zero. Such spaces are called "aspherical" and more generally, if precisely one homotopy group is nonzero, "Eilenberg MacLane spaces". In the aspherical case, these are denoted $$K(G,1)$$.
The general theorem is this. Pointed homotopy classes of pointed maps $$K(G,1) \to K(H,1)$$ are in bijection with the set of homomorphisms $$G \to H$$. (There is a proof of this in Hatcher's 1.B.) If you want unbased homotopy classes, you quotient by the action of $$G$$ and $$H$$ acting on this set by conjugacy.
Such a property is essentially never true if the domain or codomain are not both Eilenberg MacLane spaces $$K(G,n)$$ with the same $$n$$, unless your result is of the form "All maps are null-homotopic".
• @chikurin Thanks, updated. Fundamentally the idea is to build a homotopy by hand, inductively on a cell decomposition. Because you know the space $K(H,1)$ has no higher-dimensional homotopy groups, any map from an $n$-sphere to it necessarily factors through a map from the $(n+1)$-ball. This is enough to build the homotopy. – user98602 Nov 8 '18 at 3:09