Total Space of Classifying Space Contractible I have a question about the topological intuition that the total space $EG$ of the classifying space $BG$ is contractible.
Danny Calegari's "NOTES ON FIBER BUNDLES" on page 4 there is suggested following consideration:

For a group $G$ the total space $EG$ is homotopy equivalent to the join $G*G*...*G$.
I don't understand why this "argument" justifies the contractibility of $EG$.
For example if we take $G:=\{e, a\} \cong \mathbb{Z}/2$ then the join $G*G$ is homotopy equivalent to $S^1$ which is hot contractible. 

Where is the error in my reasonings? Is there an intuitive approach to understand/imagine why this space $EG \sim G*G*...*G$ is contractible?
(...I know there is still a long way for me to become a Picasso...)
 A: The claim is that the infinite join $G*G*G*\dots$ is contractible, so your example with $G*G$ is irrelevant.  As the argument says, you can justify this using the fact that joins increase connectivity: if $X$ is $n$-connected and $Y$ is $m$-connected then $X*Y$ is $(n+m+1)$-connected.  (See the comments at Connectedness of the join of two spaces for a sketch of one way to prove this fact.)  It follows by induction that the $n$-fold join $G*G*\dots*G$ is $(n-2)$-connected for all $n$ and so the infinite join is contractible.
Alternatively, here is an easier argument.  Note that if $X$ is any space and $Y$ is any nonempty space, then the inclusion $X\to X*Y$ is nullhomotopic (any point of $Y$ gives a cone over $X$ in $X*Y$).  The infinite join is just the colimit of the inclusions $G\to G*G\to G*G*G\to\dots$, and each of these inclusions is nullhomotopic, so the colimit is contractible.
For intuition, the example $G=\mathbb{Z}/2$ may be helpful.  In that case, $X*G$ is just the suspension of $X$, and so the $n$-fold join $G*G*\dots *G$ is just $S^{n-1}$, and in the limit you get the infinite sphere $S^\infty$ which is contractible.
