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?
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
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 $\alpha_{X_1}\circ F(f_1)(x_0)$.