Surjective inclusions in Van Kampen's Theorem Let $X = U \cup V$ be an open cover. Assume $U,V$ and $U \cap V$ path-connected. We have the inclusions $u : U \cap V \to U$ and $v : U \cap V \to V$ with induced maps $u_* : \pi_1(U \cap V) \to \pi_1 U$ and $v_* : \pi_1 (U \cap V) \to \pi_1 V$. From Van Kampen's Theorem $\pi_1X \cong \pi_1U \ast \pi_1V/N(K)$, where $N(K)$ is the normal closure of $K = \langle u_*(g)v_*(g^{-1})| g \in \pi_1(U \cap V)\rangle$. My Question is:
If $u_*$ and $v_*$ are surjective, is it correct that $$\pi_1X \cong \pi_1(U \cap V)/N_uN_v,$$
where $N_u$ and $N_v$ denote the kernels of $u_*$ and $v_*$ respectively.
From a paper I am reading I understand this to be correct following from van Kampen's theorem and sort of well known. I failed searching the literature and using my bare hands the calculations became too messy very soon.
 A: Yes, this is correct.   This is purely group theory, so allow me to rephrase everything solely in group language.
I'll going to set $Y = \pi_1(U\cap V)$.  I'll write $U$ for $\pi_1(U)$ and likewise $V$ will denote $\pi_1(V)$.
We have surjective homomorphisms $p_u:Y\rightarrow U$ and $p_v:Y\rightarrow V$.  I will, like you, use $N_u$ and $N_v$ for the kernels of these two maps.  I will use $N$ to the smallest normal subgroup of $U\ast V$ which contains the set $\{p_u(y) p_v(y^{-1}): y\in Y\}$; so my $N$ is your $N(K)$.
With all this notation, our goal is to prove that $Y/(N_U N_V) \cong (U\ast V)/N$.
We define $\phi: Y/(N_U N_V)\rightarrow (U\ast V)/N$ by $$\phi(y N_U N_V) = p_u(y)N.$$
Further, since $p_u$ is surjective, every element of $U$ has the form $p_u(y)$ for some $y\in Y$.  Likewise for $V$.  Thus, we can define $\psi:(U\ast V)/N\rightarrow Y/(N_U N_V)$ by $$\psi(p_u(y_1) p_v(y_2) p_u(y_3) p_v(y_4)... N) = y_1 y_2....  N_U N_V.$$
I claim that $\phi$ and $\psi$ are well defined and that they are inverses of each other.
Proposition 1:  The map $\phi$ is well defined.
Proof:  Let $n_u\in N_U, n_v\in N_V$.  We need to show that $\phi(y N_U N_V) = \phi(y n_u n_v N_U N_V)$.
Well, $\phi(y n_u n_v N_U N_V)$ is, by definition, $p_u(y n_u n_v)N$.  Since $p_u$ is a homomorphism, and since $u_n \in N_U = \ker p_u$, this is the same as $p_u(y) p_u(n_v)N$.
The element $p_u(n_v) p_v(n_v^{-1}) \in N$ by the definition of $N$.  But $n_v\in N_V = \ker p_v$, so $p_v(n_v^{-1})$ is the identity, so $p_u(n_v)\in N$.  In other words, $p_u(y)p_u(n_v) N = p_u(y) N = \phi(y N_u N_v)$.  Thus, $\phi$ is well defined. $\square$
Proposition 2:  The map $\psi$ is well defined.
Proof:  Here, there are two things to check.  First, that if $g\in U\ast V$ and $n\in N$, that $\psi(g N) = \psi(gn N)$.  Second, if $p_u(y_1) = p_u(y_1')$, $p_v(y_2) = p_v(y_2')$, etc, that $\psi(p_u(y_1)p_v(y_2)....N) = \psi(p_u(y_1') p_v(y_2').... N)$.
Let's knock out the first part first.  Recall the normal closure of a set is given by arbitrary finite products of conjugates of elements in the set.  So, it's enough to show that $\psi(n N)$ is the identity when $n$ is of the form $r p_u(y) p_v(y^{-1}) r^{-1}$ for arbitrary $r\in U\ast V$.
Writing $r = p_u(y_1) p_v(y_2)....$, we get $\psi(r p_u(y) p_v(y^{-1}) r^{-1} N) = y_1 y_2 ... y_n y y^{-1} (y_1 y_2... y_n)^{-1}$ which cancels down to the identity.
Now let's focus on the second issue.  If $p_u(y_1) = p_u(y_1')$, etc, then there are $n_1, n_3, n_5,... \in N_u$ and $n_2,n_4,... \in N_v$ with $y_i' = y_i n_i$.
So, we need to show that $y_1 y_2....N_U N_V = y_1 n_1 y_2 n_2 ...N_U N_V$.  Here's the idea.  Focus on the $n_1 y_2$ on the right side.  Rewriting this as $y_2 y_2^{-1} n_1 y_2$, the fact that $N_U$ is normal implies that $y_2^{-1} n_1 y_2 = n_1'$ for some $n_1'\in N_U$.  Thus, $n_1 y_2 = y_2 n_1'$.
Repeating this same argument, we can change $y_1 n_1 y_2 n_2....$ into $y_1 y_2 ..... n_1' n_2'......$.  Of course, the piece $n_1' n_2'....\in N_U N_V$, so $y_1 y_2... N_U N_V = y_1' y_2'.... N_U N_V$ so $\psi$ is well defined. $\square$
Proposition 3:  The maps $\psi$ and $\phi$ are inverses of each other.
Proof: First, $\psi(\phi(y N_u N_v)) = \psi(p_u(y) N) = y N_u N_v$.
Checking the $\phi\circ \psi$ is the identity is a bit messier.  It's enough to show that both $\phi(\psi(p_u(y)N)) = p_u(y)N$ and that $\phi(\psi(p_v(y)N)) = p_v(y)N$.
The first is easy:  $\phi(\psi(p_u(y)N)) = \phi(y N_u N_v) = p_u(y) N$.
The second gives $\phi(\psi(p_v(y)N)) = \phi( y N_u N_v) = p_u(y) N$, not $p_v(y)N$ like we wanted to see.  However, $p_u(y)p_v(y^{-1})\in N$ by definition, so $p_u(y)N = p_v(y)N$ and we are done. $\square$
A: I dont have a full answer, but I have some progress which might be of interest.
First, there are three maps $\pi_1(U\cap V)\to \pi_1(X)$: 


*

*$i_*$ corresponding to the inclusion $i:U\cap V\subset X$.

*By Van Kampen, the composition $\pi_1(U\cap V)\to \pi_1(U)\to \pi_1(U)*\pi_1(V)\to \pi_1(X)$

*Similarly for $V$


The Van Kampen theorem gives us that all 3 maps are actually the same, and that this map is surjective. Now we need to see that $N_u,N_v$ maps to $0$ and that their product is the entire kernel. The first part follows by choosing either definition 2 or 3 for the map.
A: I will present a solution that allows us to avoid using representations of elements in the kernel $K$.
Suppose $X=U \cup V$ where $U$ and $V$ satisfy your hypothesis, and the maps $u_*$ and $v_*$ and $N_u$ and $N_v$ and $K$ are as described above. 
Define a homomorphism $\varphi: \pi_1(A \cap B)/N_uN_v \rightarrow (\pi_1(U) * \pi_1(V))/K$ by 
$gN_uN_v \mapsto u_*(g)K$.
Observation 1: $u_*(g)K=v_*(g)K$ by definition of K.
Observation 2: $\varphi$ is well defined. 
Since if $gN_uN_v=g'N_uN_v$ then $g=g'n_un_v$ for some $n_u \in N_u$ and $n_v \in N_v$. Then $\varphi(gN_uN_v)=\varphi(g'n_un_vN_uN_v)=u_*(g'n_un_v)K=u_*(g')u_*(n_u)u_*(n_v)K=u_*(g')u_*(n_u)v_*(n_v)K=u_*(g')K$
Now we will to construct an inverse homomorphism for $\varphi$.
Observation 3: Since $u_*; \pi_{1}(U \cap V) \rightarrow \pi_1(U)$ is surjective, it induces an isomorphism $\bar{u}:\pi_1(U \cap V)/N_u \rightarrow \pi_1(U)$ and, similarly, $v_*$ induces an isomorphism $\bar{v}:\pi_1(U \cap V)/N_v \rightarrow \pi_1(V)$.
Note that, by definition, $\bar{u}(gN_u)=u_*(g)$ and $\bar{v}$ is defined similarly. 
Now define $j_u: \pi_1(U \cap V)/N_u \rightarrow \pi_1(U \cap V)/N_uN_v$ by $j_u(gN_u)=gN_uN_v$ and similarly define $j_v: \pi_1(U \cap V)/N_v \rightarrow \pi_1(U \cap V)/N_uN_v$
Observation 4: $j_u$ and $j_v$ are well defined. This is easy to check. 
Define $\mu=j_u \circ \bar{u}^{-1}$ and $\nu=j_v \circ \bar{v}^{-1}$. By the universal property of free products, $\mu$ and $\nu$ extend to a homomorphism
$\psi: \pi_1(U) * \pi_1(V) \rightarrow \pi_1(U \cap V)/N_uN_v$.
Observation 5: $K \subset \ker \psi$. 
This is also easy to check using the fact that $\bar{u}^{-1}(u_*(g))=g$ and similarly $\bar{v}^{-1}=(v_*(g))=g$
Since $K \subset \ker \psi$, we know that $\psi$ induces a homomorphism
$\bar{\psi}:(\pi_1(U)*\pi_1(V))/K \rightarrow \pi_{1}(U \cap V)/N_uN_v$
defined by $\bar{\psi}(gN)=\psi(g)$.
Observation 6: $\bar{\psi}$ is the inverse of $\varphi$. This follows straight from the definitions easily.
Conclude that $\varphi$ is an isomorphism.
