Im trying to understand the Universal Cover of the space $S^1\vee S^2\vee S^3$. I've been told it has the form of attaching a copy of $S^2\vee S^3$ at every integer of the real numbers (we call this $U$), where we associate the real number line as the helix over the circle.
It makes sence that this is simply connected, since its just three simply connectbut how do you properly define the map $p: U \rightarrow S^1\vee S^2\vee S^3$ so that i can prove this is indeed a covering space?
Thanks and cheers
Loosely, you map $\Bbb R\to S^1$ in the usual way and map the copies of $S^2,S^3$ to themselves.
More formally fix basepoints of $S^2,S^3$ and put: $$X:=\Bbb R\sqcup\bigsqcup_{n\in\Bbb Z}(S^2\vee S^3)$$And put $U:=X/_\sim$ where $n\in\Bbb R$ is $\sim$-related to $(p,n)\in\bigsqcup_{n\in\Bbb Z}S^2\vee S^3$, $p$ being the wedge point of $S^2\vee S^3$. Then there is a map $X\to S^1\vee S^2\vee S^3$ defined on its $\Bbb R$-component by $x\mapsto\exp(2\pi ix)$ and on its every $n$th $S^2\vee S^3$ component by the inclusion $S^2\vee S^2\hookrightarrow S^1\vee S^2\vee S^3$. Assuming we consider $S^1$ to be based at $1$ (this is for convenience's sake only, the choice of basepoint doesn't matter) this map clearly coarsens the $\sim$-relation and induces a continuous $U\to S^1\vee S^2\vee S^3$ and it is easy to check this is a covering map.
Why is $U$ simply connected? Let $\gamma$ be any loop in $U$. $\gamma$ has compact image bounded by some large integer $N$ (in the sense that its image is disjoint from the image of $X_N:=\Bbb R\setminus[-N,N]\sqcup\bigsqcup_{n>N}S^2\vee S^3$ in $U$) so if we can show $\gamma$ is contractible in the subspace $U_N$ - the image of $X\setminus X_N$ - then $\gamma$ will be contractible in $U$. $U$ is clearly path connected, so we would be done.
Oh, but $U_N$ is a CW complex and the finitely many contractible subcomplexes $[n,n+1]$, $-N\le n\le N-1$, can then be quotiented out without changing the homotopy type. Doing this one interval at a time, we see $U_N$ is homotopy equivalent to a wedge of numerous copies of $S^2$ and $S^3$. The wedge sum of simply connected space is simply connected, so we conclude $U_N$ is simply connected.