I have a question about a remark done by Martin Brandenburg on Tyler Lawson's answer in this MO discussion: https://mathoverflow.net/questions/10364/categorical-homotopy-colimits/10399#10399
The question dealt with looking for an explicite example that shows that the homotopy category of pointed spaces $\operatorname{hTop}_*$ in general has no pushouts.
Tylor Lawson wrote in his answer:
Your example (the "cokernel" of the multiplication by 2 map) also works.
Consider the diagram $S^1 \leftarrow S^1 \rightarrow D^2$ in the based homotopy category of CW-complexes, where the left-hand map is multiplication by 2. Suppose it had a pushout $X$ in the homotopy category. Then for any $Y$, $[X,Y]$ is isomorphic to the set of 2-torsion elements in $\pi_1(Y)$.
Taking $Y = S^0$, we find $X$ is connected.
Taking $Y = K(\pi,1)$, we find that $\pi_1(X)$ must be isomorphic to $\mathbb{Z}/2$. This means that there is a map from ${\mathbb{RP}^2}$ to $X$ inducing an isomorphism on $\pi_1$, and that there is a map $X \to K(\mathbb{Z}/2,1)$ that also induces an isomorphism on $\pi_1$.
Net result, we get a composite sequence of maps $\mathbb{RP}^2 \to X \to \mathbb{RP}^\infty \to \mathbb{CP}^\infty$. The final space is simply connected, so the map from $X$ would be nullhomotopic and hence so would the map from $\mathbb{RP}^2$.
However, the composite of the first two maps is an isomorphism on $\pi_1$, hence on $H_1$. Looking at induced maps on the second cohomology group $H^2$, we get the sequence of maps: $$\mathbb{Z}/2 \leftarrow H^2(X) \leftarrow \mathbb{Z}/2 \leftarrow \mathbb{Z}$$ The rightmost map is surjective, the composite of the two leftmost maps is an isomorphism by the universal coefficient theorem, and the composite of the two rightmost maps is supposed to be nullhomotopic and hence zero. Contradiction.
Now Martin Brandenburg remarked that the example in the answer shows also that the fundamental group functor $\pi_1 : hTop_* \to Grp$ has no left adjoint.
Could anybody explain how this conclusion work in detail? I not understand the argument.
Suppose it had a left adjoint $L: \mathbf{Grp}\to hTop_*$, then we would have, for any group $G$, $[\mathbb Z/2, \pi_1(K(G,1))] \cong [L(\mathbb Z/2),K(G,1)]$ as well as $[L(\mathbb Z/2), S^0]\cong *$.
So $L(\mathbb Z/2)$ would represent $2$-torsion in $\pi_1$ and be connected; which is all that's needed in Tyler's argument to get a contradiction.
Therefore, this left adjoint can't exist.