We are using the book by Munkres and the problem 1 in section 70 says:
Suppose that the homomorphism $i_{*}$ induced by inclusion $i: U \cap V \rightarrow X$ is trivial.
Show that $j_{1}$ and $j_{2}$ induce an epimorphism \begin{equation} h: (\pi_{1}(U,x_{0})/N_{1})*(\pi_{1}(V,x_{0})/N_{2}) \rightarrow \pi_{1}(X,x_{0}) \end{equation} where $N_{1}$ is the least normal subgroup of $\pi_{1}(U,x_{0})$ containing image $i_{1}$ and $N_{2}$ is the least normal subgroup of $\pi_{1}(V,x_{0})$ containing image $i_{2}$.
Maybe this is a dumb question, but I'm pretty confused by induced homomorphisms. What exactly does it mean for $j_{1}$ and $j_{2}$ to "induce an epimorphism"?
Note $\iota : U \cap V \hookrightarrow X$ factors as $U \cap V \hookrightarrow U \hookrightarrow X$, so that $\iota_{*} = j_1 \circ i_1 = 0$. Hence $j_1$ is $0$ on the image of $i_1$, and factors through $\overline{j_1} : \pi_1(U, x_0)/N_1 \to \pi_1(X, x_0)$. Similarly $j_2$ also factors through $\overline{j_2} : \pi_1(V, x_0)/N_2 \to \pi_1(X, x_0)$.
Now recall that if $A, B, C$ are groups and $f : A \to C$, $g : B \to C$ morphisms, they induce in a unique way a morphism $A * B \to C$ by the universal property of the free product.
In our situation, this gives the morphism $h : \pi_1(U, x_0)/N_1 * \pi_1(V, x_0)/N_2 \to \pi_1(X, x_0)$ obtained by combining $\overline{j_1}$ and $\overline{j_2}$. It is easy to verify this must be an epimorphism, since $\pi_1(X, x_0)$ is generated by the images of $j_1, j_2$.