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Struggling to answer this question from Lickorish's introduction to Knot Theory.

If M has a 3-manifold with a genus g Heegaard splitting, then the fundamental group of M has a presentation with g generators and g relators

I know that since the Heegaard splitting exists, then $M=X\cup Y$ for a pair of handlebodies X and Y, both of genus g, and that by Van-Kampen's Theorem, $$\pi_{1}(M)=\pi_{1}(X)_{*\pi_{1}(X\cap Y)}\pi_{1}(Y)$$ but I'm struggling to show that this shows that the fundamental group of M has a presentation with g generators and g relators.

Any help?

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    What is $\pi_1(X)$? How about $Y$? How about $X \cap Y$? – 2017-01-24
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    @MikeMiller If X has genus g then $\pi_{1}(X)=$. Similarly for Y but with different generators and relators. Also $\pi_{1}(X\cap Y)=\pi_{1}(\partial X)=\pi_{1}(\partial Y)$, so would this just be $\pi_{1}$ at the common boundary between the two handlebodies? – 2017-01-25
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    No, $X$ is not the boundary of the handlebody - it's the whole handlebody! $\pi_1(X \cap Y)$ is the fundamental group of the boundary genus $g$ surface (whose presentation you first wrote). – 2017-01-25

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First of all notice that the fundamental group of a CW complex with one 0-cell, $b$ 1-cells, and $r$ 2-cells admits a presentation with $b$ generators and $r$ relations.

Suppose that $M = U_0 \cup_\partial U_1$ with $U_0$ and $U_1$ genus $g$ handlebodies. Pick a collection of properly embedded two-disks $D_1, \dots, D_g \subset U_1$ such that $B=U_1- (D_1 \cup \dots \cup D_g)$ is a three ball (you can see that such a collection exists just by looking at the standard picture of a genus $g$ handlebody). Since the fundamental group depends only on the 2-skeleton we have that

$$ \pi_1(M)= \pi_1(M - B) = \pi_1(U_1 \cup_\partial(D_1\cup \dots \cup D_g)) . $$

Now observe that $U_1$ has the homotopy type of a bouquet of $g$ circles and consequently that $X=U_1 \cup_\partial(D_1\cup \dots \cup D_g)$ has the homotopy type of a CW complex with one 0-cell, $g$ 1-cells, and g 2-cells.