I want to show that $S^1_{(0)}\leftarrow *\to S^1_{(1)}$ has no pushout in the homotopy category without using Eilenberg–MacLane spaces.
In a first step, I want to show that if there is such a pushout, it must be homotopy equivalent to $S:=S^1_{(0)}\sqcup_0 [0,1]\sqcup_1 S^1_{(1)}$. Having this, it is easy to see that this cannot be a pushout.
In order to get the first statement, I call the pushout $X$ and construct a map $f:X\to S$ and $g:S\to X$.
- $f$ is just a representant for the universal map arising from $S^1_{(i)}\to S^1_{(i)}$.
- $g$ is constructed by from $S^1_{(i)}\to X$. Now the basepoints can be mapped to different points, but as the square commutes, they are in the same path component. Now we choose a good (see below) path $\gamma$ connecting these two. This completes the map $S\to X$.
By the universal property, it is clear that $g\circ f\simeq \mathrm{id}_X$. I want to proof that $f\circ g\simeq\mathrm{id}_S$. Of course, this depends on the choice of $\gamma$. To be precise, I need something of the following form:
$\gamma$ can be chosen such that $f\circ\gamma:[0,1]\to S$ is the inclusion of the intervall.
This last claim looks so elementary that there must be an easy proof or counterexample, but I cannot find it.