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Since the category of groups $\textbf{Grp}$ have all (co)products (which are free products and direct products) and all (co)equalizers, $\textbf{Grp}$ is bicomplete.

Note that a limit on a functor $F:\mathscr{D}\rightarrow \textbf{Grp}$ can be constructed as $\{(g_D)\in \prod F(D): Ff(g_D)=g_{D'} \text{ for all morphisms } f:D\rightarrow D'\}$ where $\mathscr{D}$ is a small category. This is a very simple construction.

I'm curious if there is a nice construction for a colimit on a functor just like a limit on a functor in $\textbf{Grp}$.

I'm curious about this since a good construction can give more information than that we can get from categorical argument. (e.g. Category theory does not tell anything about "normal form" of an element of a free product)

Define $S:=\{(F(f)(a),a):f:D\rightarrow D' \text{ is a morphism}, a\in D, D,D'\in \mathscr{D}\}$. Let $\bar S$ be the normal closure of the equivalence closure of $S$. Define $X:=\coprod F(D)/\bar S$. Would it be the most natural construction?

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    You mean the direct limit of groups? – 2017-01-27
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    No, a direct limit is a special case of a colimit. I mean the general colimit. – 2017-01-27
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    From my experience they are synonomous as they are the same universal property, only that direct limit is more for certain categories. – 2017-01-27
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    @ZelosMalum How do you construct the direct limit of groups? – 2017-01-27
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    For a collection of groups $(G_i)$ and homomorphisms $\varphi_{ij}$ we have that the direct limit is $\prod_i G_i/\sim$ where for $x_i\in G_i$ and $x_j\in G_j$ we have that $x_i\sim x_j$ if $\varphi_{ij}(x_i)=x_j$ – 2017-01-27
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    You may mean the quotient of the "free" product? And is $\sim$ the normal closure of the equivalence closure of that relation? – 2017-01-27
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    It is the general product of groups and the equivalence relation in question is the one when you consider the inclusion homomorphism from $G_i$ to $\prod G_i$ and the elements you get in the product. – 2017-01-27
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    @ZelosMalum I'm really confused now. I think it cannot be the direct sum (I think you meant the direct sum by the general direct product) since direct sum is not coproduct in the category of groups. However, the direct sum is the coproduct in the category of **abelian** groups. I just checked Lang-algebra and Serre-Trees, and Lang defines the direct limit of **abelian** groups by the way you wrote meanwhile Serre defines the direct limit of groups by using the free products (p.1). Would you give me a reference that defines the direct limit of groups in your way? – 2017-01-27
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    http://sierra.nmsu.edu/morandi/notes/DirectLimits.pdf – 2017-01-27
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    @Zelos Malum I read this before. The author defines direct limit of **abelian groups** in this note. There is a famous theorem saying that a category has all colimits iff it has all coequalizers and coproducts. Direct sum cannot imply the existence of free product(coproduct) by functoriality.. – 2017-01-27
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    I'll check it once I get home I will, my master thesis has it. – 2017-01-27
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    The reason limits are so nice is that the forgetful functor to set creates them. – 2017-01-27
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    This is because the category of groups is a so called algebraic category. – 2017-01-27
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    A colimit in grp is constructed as a quotient of a free product: just glue elements together as necessary. Have a look at the first few pages of Trees again. – 2017-01-27
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    "Most natural" is of course a kind of vague request. Is there any reason why you'd expect there to be an easier description than what you gave? It consists of (basically) two things, taking a free product and then a quotient. Any construction of a general colimit will need to contain both those elements as special cases, since those are the constructions you need for coproducts and coequalizers. I find it hard to imagine there's much room for improvement there. – 2017-01-27
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    I wanted to make sure if it is true. Thank you all! – 2017-01-27

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