I believe that $S^1\vee S^1$ is the Eilenberg-Mac Lane space $K(\mathbb{Z}\ast\mathbb{Z},1)$. One can prove this by constructing its universal cover and observing that it is contractible.


My question is this:


**Is there a simple homotopy-theoretic proof of this result that does not use connected covers?**


I have never seen one appear in any text. Repeated attemps by myself to use the usual tools (Hilton-Milnor Theorem, examine the fibre of the inclusion $S^1\vee S^1\hookrightarrow S^1\times S^1$, etc...) to prove this elementary result have met with connectivity issues or similar problems.



Is $S^1\vee S^1$ an Eilenberg-Mac Lane Space to a Homotopy Purist?
I think the following can be turned into a proof, but I haven't checked the details. 


By a result of Milnor, $\Omega (S^1 \vee S^1)$ coincides up to homotopy with $F(S^0 \vee S^0)$, the free group functor on the pointed set $S^0 \vee S^0$. By a version of the Hilton-Milnor theorem, the latter coincides up to homotopy with $F(S^0) \ast F(S^0)$, the latter being the free
product with amalgamation of simplicial free groups. But $F(S^0)$ coincides with $\Bbb Z$ up to homotopy. Hence, 
$\Omega (S^1 \vee S^1)$ coincides up to homotopy with $\Bbb Z \ast \Bbb Z$ (as a space). In particular, $\Omega (S^1 \vee S^1)$ is homotopically discrete, which gives the result you are after (in conjunction with the van Kampen theorem).


