The following are taken from $\textit{Arrows, Structures and Functors the categorical imperative}$ by Arbib and Manes
Let $\textbf{K}$ be the 4-object 9-morphism category
Digram 1
i.e. the morphisms are $id_A$, $id_B$, $id_C$, $a,b,c,d$ and $c\circ a=c\circ b.$ Show that $C=B+D$ but that the coproduct injection $c$ is not a monomorphism. Show that whenever $A=\coprod A_i$ in a category $\textbf{L}$, a sufficient condition for ${in}_j:A_j\rightarrow A$ to be a split monomorphism is that there exists a morphism from $A_j$ to $A_i$ for all $ i.$
Also, here is the definition of coproduct for a general category:
$\text{Definition:}$ A $\textbf{coproduct}$ of $(A_i\mid i\in I)$ is a family
$$(A_i \xrightarrow{\normalsize{\text{in}}_i} A\mid i\in I)$$
(of $\textbf{injections}$ which, consider in $\textbf{K}^{op}$
$$(A_i \xleftarrow{\normalsize{\text{in}}_i} <A|\in I)$$
is a product there; that is, given $f_i:A_i\rightarrow C$ there exists a unique $f:A\rightarrow C$ such that
$f\circ {in}_i=f_i$ for all $i$. We write $A=\coprod_{i\in I} A_i$ and if $I=\{1,\ldots, n\},$ $A=A_i + \ldots + A_n.$
Also the last part of the question where the authors ask the reader the reader to "show that whenever $A=\coprod A_i$ in a category $\textbf{L}$, a sufficient condition for ${in}_j:A_j\rightarrow A$ to be a split monomorphism is that there exists a morphism from $A_j$ to $A_i$ for all $ i".$ Also, I am not sure if the following exercise as a consequences of the following theorem is relevant: (taken from Comprehensive mathematics for computer scientists vol 1, by Guerino Mazzola, Gérard Milmeister, Jody Weissmann)
Theorem: (Universal property of coproduct) Given two sets $a$ and $b$ and any set $c$, the function
$$\gamma:Set(a\sqcup b,c)\xrightarrow{\cong} Set(a,c)\times Set(b,c)$$ Defined by $\gamma(u)=(u\circ {in}_a,u\circ {in}_{b})$ is a bijection.
Proof: Clearly, a map $u:a\sqcup b\rightarrow c$ is determined by its restrictions to its partitioning subsets $\{0\}\times a$ and $\{1\}\times b$, which in turn is equivalent to the pair $u\circ {in}_{a}$ and $u\circ {in}_{b}$ of maps. So $\gamma$ is injective. Conversely, if $v:a\rightarrow c$ and $w:b\rightarrow c$ are any two functions, then we define $u((0,x))=v(x)$ and $u((1,y))=w(y)$, which shows the surjectivity of $\gamma$.$\textbf{Exercise:}$ Suppose that a seq $q$. together with two functions $i_a:a\rightarrow q$ and $i_b:b\rightarrow q$ has the property that $$\gamma:Set(q,c)\xrightarrow{\cong} Set(a,c)\times Set(b,c)$$ defined by $\gamma(u)=(u\circ i_a,u\circ i_b)$ is a bijection. Show that there is a unique bijection $i:a\sqcup b\rightarrow q$ such that $i\circ {in}_a=i_a,i\circ {in}_b=i_b.$
Lastly from $\textit{Handbook of Categorical Algebra 1: Basica Category theory}$ by F Borceux, he provides a general definition for Coproduct in a general category along with associated propositions:
$\textbf{Questions:}$
I have few quick questions about the above exercise.
(1) For showing that $C=B+D$, is the following commutative diagram correct? I am not sure if I need to draw a pushout diagram in order to solve the problem. Also, not sure what role $c$ being the coequalizer is to play in $C$ is the coproduct of $B+D$.
(2) Should not the coproduct injection $c$ be a monomorphism. The reason for being such is that if $c$ is a coequalizer, then it implies that $a=b?$ Basically, is there a misprint to this part of the question.
(3) I added the relevant information about $A=\coprod A_J$ and ${in}_j:A_j\rightarrow A$ to the second commutative diagram above, but where should a morphism from $A_j$ to $A_i$ for all $i$ be placed in the commutative diagram below.
Thank you in advance
$(1)$:
Usually $B+D$ is notation for $B\sqcup D$, so your diagram doesn't make sense in that respect. Otherwise, your diagram is correct, but probably confusing for you. All you have to do is check that $(c:B\to C,d:D\to C,C)$ is a coproduct for $B$ and $D$ - so just check the universal property. In my experience, it's rarely easier or helpful to check: "Is the canonical arrow $B\sqcup D\to C$ an isomorphism?" and it's especially unhelpful here since a priori you don't even know that $B\sqcup D$ exists.
The universal property involves arrows out of $C$. Are there any?
$(2)$:
The whole point of this exercise is to point out coproduct inclusions don't have to be monomorphisms. It is true that $c$ is a coequaliser, but they didn't actually say that in the text. Just in case, I think I should ask you (or you should ask yourself) to check $c$ is a coequaliser in detail i.e. you should check the universal property is satisfied.
That being said, coequalisers are not guaranteed to be monomorphisms. In most of the usual categories such as $\mathsf{Set},\mathsf{Ab},\mathsf{Top}$ etc. coequalisers are hardly ever monomorphisms (monic(s)). They are epimorphisms (epic(s)), but 'epis' aren't guaranteed to be 'monos'. It is not true that $a=b$, and there is no misprint at all.
What is the definition of monomorphism? Does it apply when tested against the arrow $c$?
(I think I might recommend to you in general, that exercises in category theory tend to be solved just by "unwinding the definitions" i.e. just figure out what on Earth you have to check, explicitly, and once you've done that things become a hellavalot easier).
$(3)$:
I don't know why you've drawn that diagram, it'll only confuse you. You're supposed to check this for an arbitrary category $\mathsf{L}$.
Hint: To check $\mathrm{in}_j:A_j\to A$ is a 'split mono' (get used to the abbreviations, I'd advise) we just need to find some arrow $s_j:A\to A_j$ with $s\circ\mathrm{in}_j=\mathrm{Id}_{A_j}$, by definition.
How do we get an arrow $A\to A_j$? Well, arrows out of coproducts ($A$ being a coproduct) have quite an explicit description. Let's use that to construct the $s_j$.