What is the weakest set of assumptions on a pair of spaces $X\subset M$ for which the second homotopy group $\pi_2(M,X) $ is guaranteed to be Abelian?

Naively, I expected that Abelian $\pi_1(X)$ would do the job. However, there are counterexamples. For example, this thread on math.stackexchange discusses elaborate examples with non-Abelian $\pi_2(M,X)$ and Abelian $\pi_1(X)$ by using certain Eilenberg-MacLane spaces.

A student of mine found a simple graphical argument that (if correct) suggests that if

  1. $\pi_1(X)$ is Abelian, and

  2. $X$ is homotopic to a point in $M$

then $\pi_2(M,X)$ is indeed Abelian. This already covers all the physics applications that we have in mind. Nevertheless, it left us wondering how much can one relax the two conditions to still guarantee the commutativity of the relative homotopy group.

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    $\begingroup$ The relevant portion of homotopy exact sequence of the pair is $\pi_2(X)\to \pi_2(M)\to \pi_2(M, X)\to \pi_1(X)$, so e.g.. if $ \pi_1(X)$ is abelian, and the first arrow in the sequence is onto, then the last arrow is an isomorphism, so $ \pi_2(M, X)$ is abelian. If $X$ is null-homotopic in $M$, then the first arrow is zero, so $ \pi_2(M, X)$ is an extension with abelian kernel and quotient, but this doesn't mean that $ \pi_2(M, X)$ is abelian. $\endgroup$ – Igor Belegradek Feb 27 '18 at 22:47
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    $\begingroup$ @IgorBelegradek That part of the exact sequence doesn't give it to you, but if $X$ is null-homotopic in $M$ then there is an isomorphism $\pi_n(M,X) \cong \pi_{n-1}(X) \times \pi_{n}(M)$, as groups for $n > 1$. $\endgroup$ – Tyler Lawson Feb 27 '18 at 22:55
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    $\begingroup$ @TylerLawson: Thanks! The matter is actually discussed at length in Spanier's "Algebraic topology", chapter 7, section 3, theorem 12, i.e., if $a, b$ are in $\pi_2(M,X)$, then $aba^{-1}$ is obtained by $\partial a$-action on $b$, with respect to the usual $\pi_1(X)$ action on $\pi_2(M,X)$. $\endgroup$ – Igor Belegradek Feb 27 '18 at 23:36
  • $\begingroup$ It should be helpful to consider examples. There are many examples of crossed modules $\mu: M \to P$ such that $M,P$ are abelian, but have non trivial $2$-type: the simplest has $\mu: C_2 \times C_2 \to C_4$ with twisting action of $C_4$. There is a paper in Experimental Math. by Ellis and van Luyen on crossed modules of small order, but does not list the abelian ones. . The use of classifying spaces gives topological realisations of these. $\endgroup$ – Ronnie Brown Feb 28 '18 at 12:02

That is surely false. Let Y be any space with a nonabelian second homotopy group and let M be the one point union of Y and D^2 and let X be S^1 which is homotopic to a point in D^2 and thus M.

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    $\begingroup$ For any space $Y$, $\pi_2(Y)$ is abelian (e.g. here), and therefore this argument does not work. $\endgroup$ – Arun Debray Feb 27 '18 at 22:45

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