A monomorphism of groups which is not universal? Is there an injective homomorphism of groups $f_1\colon G\longrightarrow H_1$ together with another homomorphism $f_2\colon G\longrightarrow H_2$ such that the pushout $H_2\longrightarrow H_1\coprod_G H_2$ is not injective?
Or in other words: Is there a monomorphism in the category of groups, which is not a universal monomorphism?
 A: If $f_i :G  \to H_i$ are two homomorphisms, and $H_i = \langle X_i | R_i \rangle$ are group presentations, then the pushout has the group presentation
$$H_1 \sqcup_G H_2 = \langle X_1,X_2 : R_1, R_2 , \{f_1(g)=f_2(g)\}_{g \in G} \rangle.$$
Formally, one has to express $f_1(g)$ (likewise $f_2(g)$) here in terms of the generators $X_1$ (resp. $X_2$) so that $f_1(g)=f_2(g)$ becomes, in fact, a relation between the letters of $X_1$ and $X_2$. Now look what happens for $G=\mathbb{Z}$ and $H_2=\mathbb{Z}/n\mathbb{Z}=\langle t : t^n=1 \rangle$ with $f_2$ the canonical projection. Then $f_1 : \mathbb{Z} \to H_1$ corresponds to an element $h \in H_1$ (via $h=f_1(1)$) and we have
$$H_1 \sqcup_G H_2 = \langle X_1,t : R_1, t^n=1,h=t \rangle = \langle X_1:R_1,h^n=1 \rangle = H_1 / \langle\langle h^n \rangle\rangle.$$
Now assume that $f_1$ is injective, i.e. that $h \in H_1$ has infinite order. We ask ourselves if the canonical homomorphism $H_2 \to H_1 \sqcup_G H_2$ is also injective. It identifies with
$$\mathbb{Z}/n\mathbb{Z} \to H_1/\langle \langle h^n \rangle\rangle,\,[1] \mapsto [h].$$
Hence, the question is equivalent to: Does $[h]$ have order $n$ in $H_1 / \langle\langle h^n \rangle\rangle$? A priori it is only clear that the order divides $n$. By the way, it is a common mistake to deduce the order of the generators from a given group presentation - sometimes such a group turns out to be trivial! And this leads us to a class of counterexamples, namely $H_1$ could be simple, so that $H_1 / \langle \langle h^n \rangle \rangle = 0$ cannot contain $\mathbb{Z}/n\mathbb{Z}$ for $n>1$. An example of an infinite simple group with an element of infinite order is $\mathrm{PSL}_3(\mathbb{R})$ with $$h = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$
A: Do I misunderstand? Take an embedding of a group $i: H \hookrightarrow G$ into a simple group $G$, and suppose we have a nontrivial quotient $q: H \to Q$. The pushout of a quotient $q$ along a map $i$ must be a quotient $r: G \to P$: 
$$\begin{array}{rcl}
H & \stackrel{i}{\to} & G \\
q \downarrow & & \downarrow r \\ 
Q & \underset{j}{\to} & P
\end{array}$$
There is no way $r$ could be an isomorphism, so $r$ must be a map to the trivial group. Need I say more? 
