I'm having trouble understanding part of the proof of the classical version of the Seifert-Van Kampen theorem in Munkres:
Theorem 70.2 (Seifert-Van Kampen theorem, classical version). Let $X=U\cup V$, where $U$ and $V$ are open in $X$; assume $U, V$, and $U\cap V$ are path-connected; let $x_0\in U\cap V$. Let $$ j:\pi_1(U,x_0)*\pi_1(V,x_0)\longrightarrow\pi_1(X,x_0) $$ be the homomorphism of the free product that extends the homomorphisms $j_1$ and $j_2$ induced by inclusion. Then $j$ is surjective, and its kernel is the least normal subgroup $N$ of the free product that contains all elements represented by words of the form $$ (i_1(g)^{-1},i_2(g)), $$ for $g\in \pi_1( U\cap V, x_0)$, where $i_1,i_2$ are induced by inclusion.
Proof. The fact that $\pi_1(X,x_0)$ is generated by the images of $j_1$ and $j_2$ implies that $j$ is surjective. We show that $N\subset\ker j$. Since $\ker j$ is normal, it is enough to show that $i_1(g)^{-1}i_2(g)$ belongs to $\ker j$ for each $g\in\pi_1(U\cap V,x_0)$. If $i:U\cap V\to X$ is the inclusion mapping, then $$ ji_1(g)=j_1i_1(g)=i_*(g)=j_2i_2(g)=ji_2(g). $$ Then $i_1(g)^{-1}i_2(g)$ belongs to the kernel of $j$. Let $H$ denote the group $\pi_1(U,x_0)*\pi_1(V,x_0)/N$. It follows that $j$ induces an epimorphism $$ k:H\longrightarrow\pi_1(X,x_0). $$ We show that $N$ equals $\ker j$ by showing that $k$ is injective. It suffices to show that $k$ has a left inverse......
I couldn't understand the sentence in black. If $k$ is injective then it is an isomorphism, so $$\pi_1(U,x_0)*\pi_1(V,x_0)/N\cong\pi_1(X,x_0).$$ Why can we conclude that $N=\ker j$? All we've got are $$\pi_1(U,x_0)*\pi_1(V,x_0)/\ker j\cong\pi_1(X,x_0) \qquad and \qquad N\subset\ker j.$$ But there are plenty of examples where $A,B$ are normal subgroups of a group $G$, $G/A\cong G/B$ and $A\subset B$, but $A\ne B$. For example, $G=\mathbb Z^\omega$, $A=\mathbb Z \oplus \bigoplus\limits_{i=2}^\infty 0$ and $B=\mathbb Z^2 \oplus \bigoplus\limits_{i=3}^\infty 0$. I think this must have something to do with the fact that $\ker j$ is the kernel of a homomorphism, but what additional information does being a kernel offer?
The conclusion isn’t being drawn from the fact that if $k$ is injective then it’s an isomorphism; you’re right that that wouldn’t be enough. That $k$ is injective implies $\ker j\subset N$ because if $N\not\ni a\in\ker j$, then $aN\ne N$ but $k(aN)=k(N)$.
The first part of the proof shows that $N$ is small enough that we can even define $k$, because $j$ is constant on cosets of $N$, and the second part shows that $N$ is large enough that it divides out the entire kernel, and thus all the “non-injectivity” of $j$.