Munkres van Kampen's theorem 

The above problem is in Munkres topology exercise 70.2.
I'm trying to define a map $\phi_2$. My attempt is, first define an isomorphism $\varphi:\pi_1(U\cap V)/N\to \pi_1(V)$ where $N:=ker(i_2)$. Next define $\phi_2$ by $\phi_2([f])=\phi_1(i_1(\varphi^{-1}([f])))$. Then, $\phi_1(i_1([f]N))=\phi_1(i_1([f])i_1(N))=i_1([f])i_1(N)M=i_1([f])M$. Is this valid?
 A: This looks correct to me. In the last step, you are using that $i_1(N) \subset M$ which is the crucial piece of information in this problem. The idea, less precisely, is to use
$$ \pi_1(V, x_0) \cong \pi_1(U \cap V, x_0)/N \xrightarrow{i_1} \pi_1(U, x_0)/M = H $$
where the second map is induced by $i_1 : \pi_1(U \cap V, x_0) \to \pi_1(U, x_0)$ and exists because $i_1(N) \subset M$.
This is the correct solution but I will mention that I like to think about this slightly differently. Van Kapen's theorem identifies $\pi_1(X, x_0)$ with the amalgamated free product $G_1 *_H G_2$ where $G_1 = \pi_1(U, x_0)$ and $G_2 = \pi_1(V, x_0)$ and $H = \pi_1(U \cap V, x_0)$ (this is exactly the colimit of the diagram you wrote down in the category of groups).
Now suppose $H \to G_2$ is surjective well it should be "clear" that $G_1 *_H G_2  = G_1 / \left< \ker{(H \to G_2)} \right> $ where $\left<  \ker{(H \to G_2)}\right>$ is the normal subgroup of $G_1$ generated by the image in $G_1$ of the kernel of $H \to G_1$. Why is this? Well if everything in $G_2$ is in the image of $H \to G_2$ then it can be "amalgamated away" into $G_1$ and futhermore amalgamation kills anything in $G_1$ that comes from somthing in $H$ mapping to zero under $H \to G_2$. If you are farmiliar with category theory you should be able to show that when $H \to G_2$ is surjective then $H \to G_1 \to G_1 *_H G_2$ makes $G_1 *_H G_2$ the cokernel of $H \to G_1$ giving exactly what you want.
