Covering space of $S^1 \vee S^1$? Is this a covering space of $S^1 \vee S^1$?

I'm not sure what the map from this space onto $S^1 \vee S^1$ does. What is mapped onto which $S^1$?
 A: No this is not. From my understanding of a covering space which comes almost directly from Hatcher Section 1.3 pgs 56-59, here's an attempt at an explanation.
In the graph $Y$ you have shown there is a vertex $x_0$ with 3 incoming edges and 3 outgoing edges. If $Y$ was a covering space for $X = S^1 \vee S^1$, then there is a map $p: Y \rightarrow X$ which satisfies that there exists an open cover such that every open set $U$ in the cover has a preimage $p^{-1}(U)$ which maps homeomorphically ontop $U$ by $p$. So consider the point $p(x_0) \in X$. It is contained then in some open set $U$ in the cover. So naturally $p^{-1}(U)$ maps homeomorphically onto $U$ so as $x_0 \in p^{-1}(U)$ then $x_0$ has 3 incoming edges and 3 outgoing edges then so must $p(x_0)$. As no point in $X$ has 3 incoming edges and 3 outgoing edges, then $p$ cannot exist.
So $Y$ isn't a covering space for $X$.
Hope this helps!
A: It can't be a covering space. The central point is not (homeomorphic to) a point of $S^1 \vee S^1$. But I believe this is the only problem. (It has a total of 3 incoming and 3 outgoing arrows.)
