What is wrong with this application of van Kampen's Theorem? My understanding of van Kampen's Theorem (simplified to just two neighbourhoods):
Let $X$ be a topological space and let $\{N_a, N_b\}$ be a cover of $X$ such that $N_a \cap N_b$ is path-connected (and each open set is path-connected). Then, 
$$
\pi_1X\cong \frac{\pi_1N_a*\pi_1N_b}{[i_a(\gamma)][i_b(\gamma)]^{-1}}
$$
Where $i_a$ and $i_b$ are the inclusion maps from $N_a\cap N_b$ to $N_a$ and $N_b$ respectively, and $\gamma$ is any loop in the intersection $N_a\cap N_b$ (so we quotient by the normal subgroup generated by $[i_a(\gamma)][i_b(\gamma)]^{-1}$). 
I was trying to make sure I understand it by seeing if I can "break" it: Let $N_a = N_b = X$ (where $X$ is a path-connected topological space). Then clearly $N_a$, $N_b$, and $N_a\cap N_b$ are path-connected and we may apply van Kampen. But now the intersection is just $X$, and so the inclusion maps $i_a$ and $i_b$ are just the identity maps, so applying van Kampen we have ($\gamma$ in $X$)
$$
\pi_1X\cong \frac{\pi_1N_a*\pi_1N_b}{[\gamma][\gamma]^{-1}} \cong \pi_1X*\pi_1X
$$
Clearly this is wrong, so I was wondering where my mistake is. 
 A: Your mistake is in identifying the subgroup you are quotienting out.  You quotient out by the normal subgroup generated by all elements of the form $[i_a(\gamma)][i_b(\gamma)]^{-1}$.  When you write this, $[i_a(\gamma)]$ is to be interpreted as an element of $\pi_1(N_a)$, which is then considered as an element of $\pi_1(N_a)*\pi_1(N_b)$ via the canonical inclusion map $\pi_1(N_a)\to \pi_1(N_a)*\pi_1(N_b)$.  Similarly, $[i_b(\gamma)]$ is to be interpreted as an element of $\pi_1(N_b)$, which is then considered as an element of $\pi_1(N_a)*\pi_1(N_b)$ via the canonical inclusion map $\pi_1(N_b)\to \pi_1(N_a)*\pi_1(N_b)$.
What this means is that even in the case where $N_a=N_b=X$, $[i_a(\gamma)]$ and $[i_b(\gamma)]$ are not the same element of $\pi_1(N_a)*\pi_1(N_b)=\pi_1(X)*\pi_1(X)$.  The first is the copy of $[\gamma]$ in the first factor of $\pi_1(X)*\pi_1(X)$, and the second is the copy of $[\gamma]$ in the second factor of $\pi_1(X)*\pi_1(X)$.  So, modding out a relation that says these are equal amounts to identifying the two copies of $\pi_1(X)$ inside $\pi_1(X)*\pi_1(X)$ with each other.  When you do this, the result you get is just a single $\pi_1(X)$ since the two copies have been identified, just like you want.  (Of course, this is not a rigorous proof that the quotient is $\pi_1(X)$, but it is the intuition from which you can build a proof.)
