Maclane's planarity criteria states that
A graph $G$ is planar if and only if its cycle space has a basis such that each edge of $G$ belongs to at most two elements of it.
I am reading its proof from the book Combinatorial Problems and Exercises by Lovasz. The proof is given on Page 297-299 as the solution of Problem 5.35.
One component of the proof is given as follows:
Let $G$ be a graph having circuits $C_1,\dots,C_f$ which constitute a basis of $W_G$. Further suppose that each edge of $G$ is contained in at most two of $C_1,\dots,C_f$ ($W_G$ is the cycle space of $G$). Show that either all blocks of $G$ are single circuits or there are two $C_i$'s, say $C_1$ and $C_2$ such that $C_1+C_2$ is a circuit. (Here $C_1+C_2$ refers to the operation in the cycle space.)
The proof proceeds as follows: If $C_1,\dots,C_f$ are the only circuits then these are all the blocks (an exception are $K_2$'s). Now let $K=\sum_{i=1}^r C_i$ be a circuit such that its representation has a minimum number of terms where $C_1\cap C_2\ne\emptyset$. (Clearly at least two circuits $C_i,C_j$ must share an edge for else $K$ cannot be a circuit. Without loss of generality we may take these to be $C_1,C_2$.)
Then the author goes on to show that
$K\subset C_1+C_2$ --- (2).
I have no problems with this part.
To prove the reverse inclusion (and thereby conclude the proof) the author then proceeds by contradiction. If $e$ is an edge of $C_1+C_2$ there is no loss of generality in assuming that $e\not\in K\cup C_2$. Then as $e$ is ultimately annihilated in $K=\sum_{i=1}^r C_i$ there must be a $C_k$ for $3\le k\le r$ such that $e\in C_k$ so that $C_1\cap C_k$ share an edge. Now by the method previously applied for $C_1,C_2$ one concludes that
$K\subset C_1+C_k$. --- (3)
This is followed by the statement:
But (2) and (3) imply that an edge of $K$ not in $C_1$ must belong to both of $C_2$ and $C_k$ and thus it cannot occur in $\sum_{i=1}^r C_i$, a contradiction.
I don't understand this statement. If $f\in K,C_2,C_k$ with $f\not\in C_1$ it is possible to still have $f$ in $\sum_{i=1}^r C_i$ if it occurs in an odd number of $C_i$'s in the sum.
Can someone kindly resolve my doubt and explain this statement.
You are given the assumption that each edge of $G$ is contained in at most two of the basis circuits $C_1, \dotsc, C_f$. So, if an edge is in both $C_2$ and $C_k$ with $k \neq 2$, then it cannot occur in any other summand and so it cannot occur in $K$.