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.
Best Answer
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.)