# pushout as a special case of a colimit

I want to understand to understand the concept of a colimit. At first, the definition:

Def. colimit: Let $F:C\to D$ be a functor on catgeories $C$ and $D$. An object $K$ in $D$ together with morphisms $i_X:F(X)\to K$ for each object $X$ in $D$ is a colimit of $F$, if:

(i)compability: For each morphism $f:X\to Y$ in $C$ it is $i_Y\circ F(f)=i_X$.

(ii)universality: For each object $A$ in $D$ with morphisms $\alpha_X:F(X)\to A$ such that $\alpha_Y \circ F(f)=\alpha_X$ for all morphisms $f:X\to Y$ in $C$, there exists an unique morphism $\alpha:K\to A$ such that $\alpha_X=\alpha\circ i_X$ for every object $X$ in $C$.

Example:Let $\Lambda$ be an index set and $F:C\to TOP$ a functor on a small category $C$ into the category TOP (with objects=topological spaces and morphisms=continuous maps). Let $\coprod\limits_{\lambda\in\Lambda}F(\lambda)$ be the disjoint union of topological spaces (which is a coproduct in the catgery TOP). We consider the equivalence relation $\sim$ on this disjoint union, which is generated by: $x_{\mu_1}\in F(\mu_1)$, $x_{\mu_2}\in F(\mu_2)$, if there is a morphism $f:\mu_1\to\mu_2$ such that $F(f)(\mu_1)=\mu_2$.

Then $$\operatorname{colim} F:=\coprod\limits_{\lambda\in\Lambda}F(\lambda)/\sim$$ together with morphisms $i_{\mu}:F(\mu)\to \operatorname{colim} F$ for $\mu\in\Lambda$ (which is the composition of the inclusion $F(\mu)\hookrightarrow \coprod\limits_{\lambda\in\Lambda}F(\lambda)$ and the canonical quotient map ) is a colimit in TOP.

So far, so good.

Now, I'm stuck on an explicit example (this is a discussion that a pushout in TOP in is special colimit): Let $C$ be the category with the three objects $X_0, X_1, X_2$ and the morphisms are the identity morphisms and $f_1:X_0\to X_1$ and $f_2:X_0\to X_2$. Then a colimit of a functor $F:C\to TOP$ is given by $$K=X_1\sqcup X_2/\sim$$ with $\sim$ defined as follows: $x_1\in X_1,x_2\in X_2$, then $x_1\sim x_2$ if there exists $x_0\in X_0$ such that $f_1(x_0)=x_1$ ans $f_2(x_0)=x_2$.

Questions: I'm not sure why $K=X_1\sqcup X_2/\sim$ looks like $\operatorname{colim} F:=\coprod\limits_{\lambda\in\Lambda}F(\lambda)/\sim$, since I'm stuck on how $F$ looks like. Is here $C$ the category with object-set=$\Lambda =\{0,1,2\}$ and morphisms: the identity morphisms and $f_1:0\to 1$, $f_2:0\to 2$? Then, $F(0)=X_0$, $F(1)=X_1$, $F(2)=X_2$?

And I don't understand

1.why the relation: $x_1\in X_1,x_2\in X_2$, then $x_1\sim x_2$ if there exists $x_0\in X_0$ such that $f_1(x_0)=x_1$ ans $f_2(x_0)=x_2$

is a special case as $x_{\mu_1}\in F(\mu_1)$, $x_{\mu_2}\in F(\mu_2)$, if there is a morphism $f:\mu_1\to\mu_2$ such that $F(f)(\mu_1)=\mu_2$. For me, they look completely different.

And 2. Why we take the disjoint union $X_1\sqcup X_2$ instead of $X_0\sqcup X_1\sqcup X_2$.

Can you help me to understand this example?

• The answer is that the constructions aren't identical, but they both have the universal property of the colimit, so they give homeomorphic spaces. It would be good to try to justify this yourself, but I'm happy to come by later to explain properly. Commented Dec 27, 2016 at 16:05

An object $A$ in $D$ together with morphisms $F(X)\xrightarrow{\alpha_X} A$ such that $\alpha_Y\circ F(f)=\alpha_X$ for all morphisms $X\xrightarrow{f}Y$ in $C$ is a cocone for the functor $C\xrightarrow{F}D$. Then

1. The compatibility condition on a colimit consisting of an object $K$ and morphism $F(X)\xrightarrow{i_X}K$ then says that the colimit it a cocone for the functor $C\xrightarrow{F}D$.
2. The universality condition on a colimit says that it is an initial cocone in the sense that every other cocone $F(X)\xrightarrow{\alpha_X}A$ for $C\xrightarrow{F}D$ factors as $\alpha_X=\alpha\circ i_X$ for a unique morphism $K\xrightarrow{\alpha}A$ in $D$.

Now, any cocone $F(X)\xrightarrow{\alpha_X}A$ for $C\xrightarrow{F}D$ is vacuously a cocone for $\Lambda\hookrightarrow C\xrightarrow{F}D$ where $\Lambda=\mathrm{Ob}(C)$ is the category with the same objects as $C$ but only identity morphisms. The disjoint union $\bigsqcup_{\lambda\in\Lambda}F(\lambda)$, together with inclusion morphisms $F(\lambda)\xrightarrow{i'_\lambda}\bigsqcup_{\lambda\in\Lambda}F(\lambda)$ is the colimit of $\Lambda\hookrightarrow C\xrightarrow{F}D$. Consequently, every cocone $F(X)\xrightarrow{\alpha_X}A$ for $C\xrightarrow{F}D$ factors as $\alpha_\lambda=\alpha'\circ i'_\lambda$ for a unique morphism $\bigsqcup_{\lambda\in\Lambda}F(\lambda)\xrightarrow{\alpha'}A$.

However, arbitrary morphisms $\bigsqcup_{\lambda\in\Lambda}F(\lambda)\xrightarrow{\alpha'}A$ determine cocones for $F(X)\xrightarrow{\alpha'\circ i'_X}A$ only for $\Lambda\hookrightarrow C\xrightarrow{F} D$, not for $C\xrightarrow{F}D$. The cocones for the latter functor require compatibility conditions, which cocones for the former do not have to satisfy. It turns out that the cocone determined by a morphism $\bigsqcup_{\lambda\in\Lambda}F(\lambda)\xrightarrow{\alpha'}A$ satisfies the compatiblity conditions for $C\xrightarrow{F}D$ if and only if it is itself compatible with the equivalence relation $\sim$ on $\bigsqcup_{\lambda\in\Lambda}F(\lambda)$ generated by $x_{\mu_1}\in F(\mu_1)\sim x_{\mu_2}\in F(\mu_2)$ if there exists a morphism $\mu_1\xrightarrow{f}\mu_2$ in $C$ so that $Ff(x_{\mu_1})=x_{\mu_2}$. Indeed, such compatiblity is exactly the requirement that $\alpha'\circ i'_{\mu_2}\circ F(f)(x_{\mu_1})=\alpha'\circ i'_{\mu_1}(x_{\mu_1})$, i.e. $\alpha'\circ i'_{\mu_2}\circ F(f)=\alpha'\circ i'_{\mu_1}$ for all $\mu_1\xrightarrow{f}\mu_2$ in $C$.

Since by definition the quotient $\bigsqcup_{\lambda\in\Lambda}F(\lambda)\twoheadrightarrow\bigsqcup_{\lambda\in\Lambda}F(\lambda)/_\sim$ has the property that $\bigsqcup_{\lambda\in\Lambda}F(\lambda)\xrightarrow{\alpha'}A$ factors (uniquely) as $\bigsqcup_{\lambda\in\Lambda}F(\lambda)\twoheadrightarrow\bigsqcup_{\lambda\in\Lambda}F(\lambda)/_\sim\xrightarrow{\alpha}A$ if and only if $\alpha'$ is compatible with the equivalence relation, it follows that cocones $F(X)\xrightarrow{\alpha_X}A$ factor uniquely as $F(X)\xrightarrow{i_X}\bigsqcup_{\lambda\in\Lambda}F(\lambda)/_\sim\xrightarrow{\alpha}A$ where $F(X)\xrightarrow{i_X}\bigsqcup_{\lambda\in\Lambda}F(\lambda)/_\sim$ are the composites $F(X)\xrightarrow{i'_X}\bigsqcup_{\lambda\in\Lambda}F(\lambda)\twoheadrightarrow \bigsqcup_{\lambda\in\Lambda}F(\lambda)/_\sim$.

In the case of a pushout, $C$ is the category $1\xleftarrow{f_1}0\xrightarrow{f_2}2$. Instead of taking $\Lambda=\mathrm{Ob}(C)=\{0,1,2\}$, we can take $\Lambda'=\{1,2\}$.

Then cocones $F(X)\xrightarrow{\alpha_X}A$ for $C\xrightarrow{F}D$ again determine cocones for $\Lambda'\hookrightarrow C\xrightarrow{F}D$, hence again determine morphisms $X_1\sqcup X_2\xrightarrow{\alpha'}A$, explicitly such that $\alpha_{X_1}=\alpha'\circ i'_{X_1}$ and $\alpha_{X_2}=\alpha'\circ i'_{X_2}$.

Notice, however, that unlike with $X_0\sqcup X_1\sqcup X_2$, we have to check that two cocones for $C\xrightarrow{F}D$ cannot determine the same morphism $X_1\sqcup X_2\xrightarrow{\alpha'}A$. This is indeed the case because $\alpha_{X_0}(x_0)=\alpha_{X_1}\circ F(f_1)(x_0)=\alpha_{X_1}(x_1)$.

For the compatibility condition, again not every morphism $X_1\sqcup X_2\xrightarrow{\alpha'}A$ determines a cocone for $C\xrightarrow{F}D$. However, a morphism $X_1\sqcup X_2\xrightarrow{\alpha'}A$ is compatible with the equivalence relation $x_1\in X_1\sim x_2\in X_2$ if there exists $x_0\in X_0$ with $F(f_1)(x_0)=x_1$, $F(f_2)(x_0)=x_2$ if and only if $\alpha'\circ i_{X_1}\circ F(f_1)(x_0)=\alpha'\circ i_{X_2} F(f_2)(x_0)$. Furthermore, it is only for these morphisms that you can define $\alpha_{X_0}(x_0)$ as $\alpha_{X_1}\circ F(f_1)(x_0)$.

• thank you very much, and sorry for the late reaction (I needed a while to understand this, but now I understand). Very helpful!
– Ryan
Commented Jan 10, 2017 at 10:40