Showing that every finitely presented group has a $4$-manifold with it as its fundamental group 
Wikipedia: For any finitely presented group it is easy to construct a (smooth) compact 4-manifold with it as its fundamental group.

Question: How do we do this?

EDIT: Below is a proof sketch found elsewhere with some targeted questions of my own bolded:



*

*Let $F = \left\langle S \mid R \right\rangle$ s.t. $|S| \in \mathbb{N}$.

*Let $X$ be a $4$-manifold.

*Represent each relation in $R$ by a loop in $X$.  
What exactly does this mean?

*Assume that the loop is a smooth, simple closed curve.  

*Replace the loop by a tubular neighborhood $N$, which is homeomorphic to $S^1 \times D^3$.  

*The boundary of $N$ is homeomorphic to $S^1 \times S^2$.  

*Delete the interior of $N$.  

*Now $S^1 \times S^2$ is also the boundary of $D^2 \times S^2$, so take a copy of $D^2 \times S^2$ and use it to fill in the whole by identifying on the boundary.
How is this identification procedure carried out?

*Now use Seifert Van-Kampen to show that you have killed off the generator.
I'm not exactly sure what is meant here, but I think once I understand the above bolded this will make more sense.
 A: Start with the closed $4$-manifold $X = (S^3 \times S^1) \# \cdots \# (S^3 \times S^1)$, the connected sum of $|S|$ copies of $S^3 \times S^1$. Using Seifert-van Kampen, you can check that $\pi_1(X) \cong \langle a_1, \dots, a_{|S|}\rangle$, where $a_i$ is represented by the $S^1$-factor of the $i^\text{th}$ $S^3 \times S^1$ summand of $X$.
Each relation can be represented by a smooth loop in $X$. For example, the relation $a_2 a_1^2 a_3^{-1}$ is represented by the loop in $X$ that first goes around the $S^1$-factor in the $2^\text{nd}$ $S^3 \times S^1$-summand, then goes around the $S^1$-factor in the $1^\text{st}$ $S^3 \times S^1$-summand twice, and finally goes around the $S^1$-factor in the $3^\text{rd}$ $S^3 \times S^1$ factor in reverse (we need to have an orientation chosen for each $a_i$). Denote the loop in $X$ corresponding the the $j^\text{th}$ relation by $b_j$.
Each loop $b_j$ has a tubular neighborhood $N_j$, which is a copy of $S^1 \times D^3$ embedded in $X$. The boundary of $N_j$ is homeomorphic to $S^1 \times S^2$. Note that $D^2 \times S^2$ also has boundary $S^1 \times S^2$. Hence we can cut out $N_j$ from $X$ and insert a copy of $D^2 \times S^2$ in its place by attaching it to the "empty" $S^1 \times S^2$ left behind in $X \setminus \text{int}(N_j)$.
The only step left is to check that removing $N_j$ and replacing it with a copy of $D^2 \times S^2$ has the effect of killing $b_j$. Let $X_j$ denote the manifold obtained from $X$ by this process. We have that
$$X_j = (X \setminus \text{int}(N_j)) \cup_{S^1 \times S^2} (D^2 \times S^2).$$
Now
\begin{align*}
\pi_1(X \setminus \text{int}(N_j)) & = \langle a_1, \dots, a_{|S|} \rangle, \\
\pi_1(D^2 \times S^2) & = 0, \\
\pi_1(S^1 \times S^2) & = \langle c \rangle,
\end{align*}
where $c$ is represented by $S^1 \times \{\text{pt}\}$ in $S^1 \times S^2$. Note that the image of $c$ in $X_j$ is precisely the curve $b_j$. Therefore by Seifert-van Kampen we have that
$$\pi_1(X_j) = \langle a_1, \dots, a_{|S|} \mid b_j \rangle.$$
If $X'$ denotes the result of doing this surgery of all $b_j$'s, we have that
\begin{align*}
\pi_1(X') & = \langle a_1, \dots, a_{|S|} \mid b_1, \dots, b_{|R|} \rangle \\
 & = \langle S \mid R \rangle.
\end{align*}
