# Does the proof that the homotopy groups are abelian in Switzer's book only work for hausdorff spaces?

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?

• 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 May 22 at 8:05

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.
• 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. May 22 at 13:39
• @BiancaOliveira Yes, $Z$ has to be locally compact Hausdorff. I edited my answer. And $\bar f$ is continuous simply because $E$ is a bijection. May 22 at 14:59