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The proof that the homotopy groups of a pointed space, $\pi_n(X,x_0)$, are abelian for $n\geq 2$, given by Robert P. Switzer in his book "Algebraic Topology - Homology and Homotopy", relies on the proposition 2.23 which states that the adjoint correspondence $$A:[SX,\ast;Y,y_0]\to[X,x_0;\Omega Y,\omega_0]$$ is a isomorphism of groups.
However, the definition of $A$ is given in Theorem 2.5, as follows.

Theorem. If $(X,x_0),\ (Y,y_0),\ (Z,z_0)$ are pointed topological spaces, such that $(X,x_0),\ (Y,y_0)$ are Hausdorff and $(Z,z_0)$ is locally compact, then there is a natural correspondence $$A:[Z\land X,\ast;Y,y_0]\to[X,x_0;(Y,y_0)^{(Z,z_0)},f_0]$$ defined by $A[f]=[\hat{f}]$, where if $f:Z\land X\to Y$ is a map then $\hat{f}:X\to Y^Z$ given by $(\hat{f}(x))(z)=f[z,x]$.

The proof uses the fact that $X$ is Hausdorff to show that $A$ is a bijection. My conclusion was that the proposition 2.23 assumes that $X$ is Hausdorff. But isn't the result that homotopy groups are abelian for $n\geq 2$ true regardless of the space $X$? Is the proof in Switzer's book incomplete?

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  • $\begingroup$ Part of me wonders if some topologists just have an overt hate of non-Hausdorff spaces. It annoys the completionist in me to not know what happens in those cases $\endgroup$ May 22 at 8:05

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You are right, as it is presented by Switzer, the map $A$ is known to be an isomorphism of groups only for Hausdorff $X$.

The problem goes back to 0.11 where Switzer states that the exponential function $E : Y^{X \times Z} \to (Y^Z)^X$ is a homeomorphism provided $X$ is Hausdorff and $Z$ is locally compact Hausdorff. This is true, and one cannot drop the assumption that $X$ is Hausdorff.

BUT: For Proposition 2.23 we do not need to know that $E$ is a homeomorphism, it suffices to know that it is a bijection. And that is true without the assumption that $X$ is Hausdorff, we only need to assume that $Z$ is locally compact Hausdorff. To see this, you have to delve into the proof of 0.11.

Once we have the above variantof 0.11, we can prove adequate variants of Proposition 0.13 and Theorem 2.5.

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    $\begingroup$ This answer does not deserve a downvote. $\endgroup$ May 22 at 10:39
  • $\begingroup$ Thank you! I will try to prove that $A$ is a group isomorphism using only that $E$ is a bijection. My fear is that i cant overpass the problem that $\overline{f}=E^{-1}f'$(as in the proof of Theorem 2.5) may not be continuous, i.e. $f[z,x]=\overline{f}(z,x)$ is not in $[Z\land X,\ast;Y,y_0]$. Probably i just have to find another way. Also, i assume you meant $Z$ is locally compact Hausdorff. $\endgroup$ May 22 at 13:39
  • $\begingroup$ @BiancaOliveira Yes, $Z$ has to be locally compact Hausdorff. I edited my answer. And $\bar f$ is continuous simply because $E$ is a bijection. $\endgroup$
    – Paul Frost
    May 22 at 14:59

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