In Edmond's Blossom algorithm, why do we only consider even-level nodes? I'm trying to learn Edmond's Blossom algorithm for finding a maximum matching in a general graph.
According to Wikipedia:

Given $G = (V, E)$ and a matching $M$ of $G$, a blossom $B$ is a cycle in $G$ consisting of $2k + 1$ edges of which exactly $k$ belong to $M$, and where one of the vertices $v$ of the cycle (the base) is such that there exists an alternating path of even length (the stem) from $v$ to an exposed vertex $w$.

Then it says that in order to find a blossom, you search the graph marking even-level nodes as "outer" and odd-level nodes as "inner", and:

If we end up with two adjacent vertices labeled as outer "o" then we have an odd-length cycle and hence a blossom.

I don't understand why we only consider outer (even-level) nodes. In a graph with edges $(1, 2), (2, 3), (3, 4), (3, 5), (4, 5)$, with the matching $(2, 3), (4, 5)$, node 1 is on level 0, node 2 is on level 1, node 3 is on level 2, nodes 4 and 5 are on level 3. It seems to me like the cycle (3, 4, 5) respects the definition for an odd alternating cycle, yet the algorithm will not consider it because 4 and 5 are on an odd level.
I thought it might be a wikipedia issue, but I've seen a couple of implementations, including a paper from a course at Standford, and they all consider only even-level nodes. What am I missing?
Later edit: to further illustrate my point, in the two graphs in the image below, both $(3, 4, 5)$ (in the left graph) and $(3, 4, 5, 6, 7)$ (in the right graph), are odd alternating cycles. However, the horizontal edge $(4, 5)$ is on a different parity level than the horizontal edge $(6, 7)$. Therefore, no matter how you look at it, one of them should not be considered a cycle during the algorithm, right? Which does not make much sense.

 A: The even/odd level, or equivalently the labeling of vertice as "inner" or "outer", is done with respect to the forest $F$ consisting of the portion of the graph we've already explored. In your example, before we've discovered edge $35$, $F$ will consist of the path from $1$ to $5$, so vertices $3$ and $5$ will both be at even levels: $2$ and $4$. Therefore, edge $35$ connects two vertices at even levels, and so the cycle $(3,4,5)$ is correctly reported as a blossom.
In more detail, here is what will happen:


*

*We start out with $F$ consisting only of vertex $1$. Edges $23$ and $45$ are marked.

*Vertex $1$ is an outer vertex of $F$, and there is an unmarked edge $12$. It goes to a vertex $2$ outside $F$, which is matched to $3$. We add all this to $F$; $F$ now has vertices $1,2,3$ and edges $12,23$. We mark the edge $12$.

*Vertex $1$ is an outer vertex of $F$, and there are no unmarked edges out of it, so we mark vertex $1$.

*Vertex $3$ is an outer vertex of $F$, and there is an unmarked edge $34$. It goes to a vertex $4$ outside $F$, which is matched to $5$. We add all this to $F$; $F$ now has vertices $1,2,3,4,5$ and edges $12,23,34,45$. We mark the edge $34$.

*Vertex $3$ is an outer vertex of $F$, and there is an unmarked edge $35$. It goes to another outer vertex of $F$, $5$ (it's at level $4$). These vertices have the same root, $1$, so we contract the blossom formed by edge $35$ and the unique path from $3$ to $5$ in $F$.

