Pullback of $\mathbb{Z} \xrightarrow{m} \mathbb{Z} \xleftarrow{n} \mathbb{Z}$ in $\operatorname{Ab}$? I'm trying to understand why the pullback of 
$$
\mathbb{Z} \xrightarrow{m} \mathbb{Z} \xleftarrow{n} \mathbb{Z} 
$$
in $\operatorname{Ab}$ with $m,n \neq 0$ is
\begin{array}
\ \mathbb{Z} \xrightarrow{b} \mathbb{Z} \\
\downarrow^{a}  \quad  \downarrow^m \\ \tag{1}
\mathbb{Z} \xrightarrow{n} \mathbb{Z}
\end{array} 
with $an = bm = \operatorname{lcm}(n,m)$. I understand that, for starters, for an abelian group to be the summit of such a cone, it must be free and so it has to be of the form $\mathbb{Z}^d$. 
I also see that, if it were $d = 1$ and $f,g : \mathbb{Z} \rightarrow \mathbb{Z}$ with $f \simeq c \in \mathbb{Z}, g \simeq d \in \mathbb{Z}$  such that $mc = nd$, then both $mc$ and $nd$ are divided by $\operatorname{lcm}(n,m)$ and so we have that $a = \frac{\operatorname{lcm}(n,m)}{n} | d$ and $b = \frac{\operatorname{lcm}(n,m)}{m} | c$. Thus $f$ and $g$ factor through $a$ and $b$ as desired. 
What I am missing is, how do we see that cones from $\mathbb{Z}^d$ for $d > 1$ factor trough $(1)$?
 A: If $P$ is the pullback of that cospan, then you have an exact sequence
$$0 \to P \to \mathbb{Z}^2 \xrightarrow{(m, -n)} \mathbb{Z} $$
so $P$ has to be a subgroup of $\mathbb{Z}^2$. Since $\mathbb{Z}^2$ is free, $P$ must also be free, as you observe.
Taking the tensor product with $\mathbb{Q}$ (or with any field of characteristic zero) is exact, so we also have an exact sequence
$$0 \to P \otimes \mathbb{Q} \to \mathbb{Q}^2 \xrightarrow{(m, -n)} \mathbb{Q} $$
Linear algebra tells us that $P \otimes \mathbb{Q}$ is a one-dimensional $\mathbb{Q}$-vector space. Since $P$ is free, that means $P = \mathbb{Z}^1$.

Another way to see this is to recall that pullbacks of monic maps are monic:
$$ \require{AMScd} \begin{CD}
P @>>> \mathbb{Z}
\\ @VVV @VVmV
\\ \mathbb{Z} @>n>> \mathbb{Z} \end{CD} $$
Since $\mathbb{Z} \xrightarrow{n} \mathbb{Z}$ is monic, the top map $P \to \mathbb{Z}$ must also be monic: $P$ must be (isomorphic to) a subgroup of $\mathbb{Z}$.
In regard to identifying the two maps out of $P$, I find it useful to apply an isomorphism to another pullback square whose maps are inclusions:
$$ \begin{CD}
P @>>> m \mathbb{Z}
\\ @VVV @VVV
\\ n\mathbb{Z} @>>> \mathbb{Z} \end{CD} $$
In the category of abelian groups, pullback of subgroups is given by taking intersections, so we see $P \cong \operatorname{lcm}(m,n) \mathbb{Z}$, and the maps out of $P$ are the canonical inclusions.
A: The pullback is the kernel of $\Bbb Z\oplus\Bbb Z\to\Bbb Z$ defined by
$(x,y)\mapsto mx-ny$. These are parametrised by $(x,y)=(az,bz)$
so the pullback group is $\Bbb Z$ and the maps from it take $1$ to $a$
and $1$ to $b$.
A: We should have $H:=\{(x,y) \in \mathbb Z \times \mathbb Z  \mid nx=my\}$ with maps restricting canonical projections. Note that the total space is the kernel of $(x,y) \mapsto nx-my$.
Now you want to define an isomorphism $\phi:\mathbb Z \to H$ by $1 \mapsto (a,b)$.
Note that $\mathrm{Im}(\phi) \subset H$ since $an-bm=0$ by definition.
Moreover, note that if $nx=my$, then there exists some $z$ so that $(az,bz)=(x,y)$ since   if $nx=my=d$, then $lcm(n,m) \mid d$.
Injectivity is evident, so we have an isomorphism.
A: Let's look first at the situation for $d = 1$ again and look at the image of the generator $1$: 
$\require{AMScd}$
\begin{CD}
\mathbb{Z}^1 @>x>> \mathbb{Z} \\
@VyVV      @VVmV \\
\mathbb{Z} @>n>> \mathbb{Z}
\end{CD}
Through the correct reasononig you've proposed, we find out, that $x = kb$, $y = ka$, hence there is the factorization $\mathbb{Z}^1 \stackrel{k}{\rightarrow} \mathbb{Z}$. 
For $d>1$ one can repeat this for all $d$ generators separately (it is a free group), because in the following situation:
\begin{CD}
\mathbb{Z}^d @>f>> \mathbb{Z} \\
@VgVV      @VVmV \\
\mathbb{Z} @>n>> \mathbb{Z}
\end{CD}
for each generator $x_i, i\leq d$ the equality $mf(x_i) = n g(x_i)$ holds, hence $f(x_i) = k_ib$, $g(x_i)=k_ia$ and define the factoring homomorphism on the generator $x_i$ to be $\varphi(x_i) = k_i$.
