I am confused about a statement about Intersection forms (for $4$-manifolds):
[...] Define the intersection form modulo 2 $${\displaystyle \cap _{M,2}:H_{2}(M;\mathbb {Z} /2\mathbb {Z} )\times H_{2}(M;\mathbb {Z} /2\mathbb {Z} )\to \mathbb {Z} /2\mathbb {Z} }$$
by the formula
$${\displaystyle a\cap _{M,2}b=|A\cap B|\mod 2}$$
This is well-defined because the intersection of a cycle and a boundary consists of an even number of points (why?) [...].
I not understand why an intersection of a cycle and a boundary consists of an even number of points. why is it true?
For example if we take two $3$-cells aka tetrahedrons $A,B$, which share exactly one common face. their images under the boundary operator are alternating sums $\sum_{i=0}^3 (-1)^{i}dA_i$ and $\sum_{i=0}^3 (-1)^{i}dB_i$, where by $dA_i,dB_j$ we denote the $i$-th and $j$-th faces (=$2$-cells) of $A$ and $B$.
by construction $\sum_{i=0}^3 (-1)^{i}dA_i$ and $\sum_{i=0}^3 (-1)^{i}dB_i$ are boundaries and therefore also $2$-cycles. but as $A$ and $B$ share exactly one common face, there exist a $dA_i$ and $dB_j$ with $dA_i=dB_j$ and the other faces are disjunct. doesn't this contradict to the statement above that the intersection of a cycle and a boundary consists of an even number of points?