Maps between colimit objects

Question

Let $\lbrace A_i \rbrace_{i\in I}$ and $\lbrace B_j \rbrace_{j\in J}$ be two diagrams in a category $\mathcal{C}$. Let $A=\text{colim}_{i\in I} A_i$ and $B= \text{colim}_{j\in J} B_j$. Given two maps $f,g: A\rightarrow B$, assuming that the coequaliser $C$ exists, can $C$ be presented in terms of the colimit of a diagram that consists of objects $A_i$ and $B_j$?

Answer 1

It's a tempting but false conjecture! Let all the $A_i$ and $B_i$ be copies of $\mathbb{Z}$ in the category $\text{Ab}$ of abelian groups and let both of the colimits be coproducts. Then every abelian group occurs as a coequalizer of a pair of maps between such coproducts (this says exactly that every abelian group has a presentation) but not every abelian group occurs as a colimit of copies of $\mathbb{Z}$. I asked exactly this question on MathOverflow awhile ago and got a very informative negative answer.

Answer 2

Let me change notation. Let $X : \mathcal{I} \to \mathcal{C}$ and $Y : \mathcal{J} \to \mathcal{C}$ be functors, let $\mathcal{D}$ be the full subcategory of $\mathcal{C}$ spanned by the images of $X : \mathcal{I} \to \mathcal{C}$ and $Y : \mathcal{J} \to \mathcal{C}$, and let $\bar{X} = \varinjlim_\mathcal{I} h_X$ and $\bar{Y} = \varinjlim_\mathcal{J} h_Y$ in $[\mathcal{D}^\textrm{op}, \textbf{Set}]$, where we define $h_T (S) = \mathcal{C} (S, T)$. Then, $$\textstyle [\mathcal{D}^\textrm{op}, \textbf{Set}](\bar{X}, \bar{Y}) \cong \varprojlim_\mathcal{I} \varinjlim_\mathcal{J} \mathcal{C}(X, Y)$$ so a morphism $f : \bar{X} \to \bar{Y}$ can be decomposed into a family of morphisms $X (i) \to Y (j_i)$ indexed by $i \in \operatorname{ob} \mathcal{I}$, modulo a certain equivalence relation. On the other hand, even if $X$ and $Y$ have colimits in $\mathcal{C}$, at best we can only say $$\textstyle \mathcal{C} \left( \varinjlim_\mathcal{I} X, \varinjlim_\mathcal{J} Y \right) \cong \varprojlim_\mathcal{I} \mathcal{C} \left( X, \varinjlim_\mathcal{J} Y \right)$$ and although there is a canonical comparison map $$\textstyle \varprojlim_\mathcal{I} \varinjlim_\mathcal{J} \mathcal{C}(X, Y) \to \varprojlim_\mathcal{I} \mathcal{C} \left( X, \varinjlim_\mathcal{J} Y \right)$$ it is generally not a bijection.

If – this is a big if – the parallel pair $\varinjlim_\mathcal{I} X \rightrightarrows \varinjlim_\mathcal{J} Y$ in $\mathcal{C}$ you are interested in can be lifted through this comparison map, then their coequaliser is a colimit of a diagram constructed from $X$ and $Y$.

Indeed, let $f_0, f_1 : \bar{X} \to \bar{Y}$ be morphisms in $[\mathcal{D}^\textrm{op}, \textbf{Set}]$. Then we may form their coequaliser $\bar{Z}$ in $[\mathcal{D}^\textrm{op}, \textbf{Set}]$. There is a canonical diagram $R : \textbf{El} (\bar{Z}) \to \mathcal{D}$ whose colimit in $[\mathcal{D}^\textrm{op}, \textbf{Set}]$ is $\bar{Z}$. Explicitly, $\textbf{El} (\bar{Z})$ is the following category:

• The objects in $\textbf{El} (\bar{Z})$ are pairs $(D, z)$ where $D$ is an object in $\mathcal{D}$ and $z$ is an element of $\bar{Z} (C)$.
• The morphisms $(D, z) \to (D', z')$ in $\mathcal{K}$ are the morphisms $g : D \to D'$ in $\mathcal{D}$ such that $z' \cdot g = z$.
• Composition is inherited from $\mathcal{D}$.

There is an obvious projection $R : \textbf{El} (\bar{Z}) \to \mathcal{D}$ and it is straightforward to check that $\bar{Z} \cong \varinjlim_{\textbf{El} (\bar{Z})} h_R$. I claim: $\varinjlim_{\textbf{El} (\bar{Z})} R$ is the coequaliser of $\varinjlim_\mathcal{I} X \rightrightarrows \varinjlim_\mathcal{J} Y$ in $\mathcal{C}$, in the strong sense that $\varinjlim_{\textbf{El} (\bar{Z})} R$ exists in $\mathcal{C}$ if and only if the coequaliser exists in $\mathcal{C}$ and they are isomorphic.

Let $\bar{\mathcal{D}}$ be the full subcategory of $\mathcal{C}$ spanned by the objects $C$ in $\mathcal{C}$ such that, for every object $T$ in $\mathcal{C}$, the canonical map (induced by the restriction of the Yoneda embedding) $$\mathcal{C} (C, T) \to [\mathcal{D}^\textrm{op}, \textbf{Set}] (h_C, h_T)$$ is a bijection. The usual Yoneda lemma tells us $\mathcal{D} \subseteq \bar{\mathcal{D}}$, but, for any diagram $Z : \mathcal{K} \to \bar{\mathcal{D}}$, $$\textstyle \varprojlim_\mathcal{K} \mathcal{C} \left( Z, T \right) \cong \varprojlim_\mathcal{K} [\mathcal{D}^\textrm{op}, \textbf{Set}] \left( h_Z, h_T \right) \cong [\mathcal{D}^\textrm{op}, \textbf{Set}] \left( \varinjlim_\mathcal{K} h_Z, h_T \right)$$ naturally in $T$, so if $\varinjlim_\mathcal{K} Z$ exists in $\mathcal{C}$ then it is also in $\bar{\mathcal{D}}$ – in short, $\bar{\mathcal{D}}$ is closed under colimits that exist in $\mathcal{C}$. In particular, $\varinjlim_\mathcal{I} X$ and $\varinjlim_\mathcal{J} Y$ are in $\bar{\mathcal{D}}$, and if the parallel pair $\varinjlim_\mathcal{I} X \rightrightarrows \varinjlim_\mathcal{J} Y$ in $\mathcal{C}$ has a coequaliser than the coequaliser is also in $\bar{\mathcal{D}}$. Essentially, the definition of $\bar{\mathcal{D}}$ ensures that whatever can be proved about iterated colimits in $[\mathcal{D}^\textrm{op}, \textbf{Set}]$ is also true in $\mathcal{C}$, so long as all the colimits involved can be realised in $\bar{\mathcal{D}}$. Thus, if the parallel pair $\varinjlim_\mathcal{I} X \rightrightarrows \varinjlim_\mathcal{J} Y$ in $\mathcal{C}$ can be lifted to a parallel pair $\bar{X} \rightrightarrows \bar{Y}$ in $[\mathcal{D}^\textrm{op}, \textbf{Set}]$, then the coequaliser in $\mathcal{C}$ can be identified with the colimit in $\mathcal{C}$ of a diagram in $\mathcal{D}$.

It may be worth remarking that the Yoneda representation $\bar{\mathcal{D}} \to [\mathcal{D}^\textrm{op}, \textbf{Set}]$ may be neither fully faithful nor preserve colimits... but this is just rephrasing the warning in the first paragraph.