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1) If $\mathcal{A}$ consists of the Sierpinski space and $\mathcal{B}$ consists of all finite topological spaces then the concrete coreflective hull of $\mathcal{A}$ consists of all finitely generted spaces (i.e. spaces in which $x \in clA \iff x \in cl\{a\}$ for some $a \in A$). This can be seen by showing that the total sink from Sierpinski spaces is final in Top.

So let $X$ be a (finitely generated) topological space and take the total sink from Sierpinski spaces (the functions in this sink are named $f_{i}$). Let $g: X \rightarrow Y$ be a function, so that for every $f_{i}: Sierpinski \rightarrow X$, $g \circ f_{i}$ is continuous. Why is g continuous?

2) If $\mathcal{A}$ consists of one convergent sequence $A= \{0\} \cup \{\frac{1}{n} \vert n \in \mathbb{N} \} $ or if $\mathcal{B}$ consists of the full subconstruct of all metrizable topological spaces, then in both cases the concretely coreflective hull consists of all sequential spaces. Again this is seen by showing that the total sink from $A = \{0\} \cup \{\frac{1}{n} \vert n \in \mathbb{N}\}$ is final in Top.

So let $X$ be a (sequential) topological space and take the total sink from A (the functions in this sink are named $f_{i}$). Let $g: X \rightarrow Y$ be a function, so that for every $f_{i}: A \rightarrow X$, $g \circ f_{i}$ is continuous. Why is g continuous?

3) If Sierpinski spaces belong to $\mathcal{A}$, then the concretely reflective hull of $\mathcal{A}$ in Top is Top. To prove this show that for an arbitrary topological space, the source consisting of all indicatorfunctions defined on open sets is initial. Again, I'm stuck proving this argument.

As always, any help would be appreciated.

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$\newcommand{\Zobr}[3]{{#1}\colon{#2}\to{#3}}\newcommand{\Invobr}[2]{{#1}^{-1}[#2]}$ This is an equivalent characterization of finitely generated spaces: A subset $F\subseteq X$ is closed in $X$ if and only for every continuous map $\Zobr fSX$ the set $\Invobr fF$ is closed in $S$. (Where $S$ is the Sierpinski space.)

An equivalent characterization of sequential spaces: A subset $F\subseteq X$ is closed in $X$ if and only for every continuous map $\Zobr fAX$ the set $\Invobr fF$ is closed in $A$. (Where $A$ is the space you described above.) Indeed, this is just another way of saying that closed sets in $X$ are precisely the sequentially closed sets. Picture from my answer here might help to visualize this.

So in both, 1 and 2, you have the same characterization of closed sets using continuous maps from the spaces from $\mathcal A=\{A\}$ or from $\mathcal A=\{S\}$.

Now if you have a map $\Zobr gXY$, to show that $g$ is continuous you only have to show that inverse image of each closed set is closed. If $G\subseteq X$ closed then each $\Invobr {g\circ f_i}G=\Invobr{f_i}{\Invobr gG}$ is closed. But, according to above characterization, this is equivalent to saying that $\Invobr gG$ is closed.


I don't think the third thing is true, since the reflective hull of $S$ in $\mathbf{Top}$ is the subcategory $\mathbf{Top}_0$ of all $T_0$-spaces.


Great reference for reflective and coreflective subcategories of the category of topological spaces is the book H. Herrlich: Topologische Reflexionen und Coreflexionen, Lecture Notes in Mathematics 78, Springer. But basic results can be found in many books about category theory, e.g. in Adamek-Herrlich-Strecker.

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Thanks for your help! The course in categorical topology i follow is indeed based on the book about category theory by Adamek-Herrlich-Strecker. Thank you for the extra reference you gave me! It is indeed a very good one! During the lessons my professor told me the answer to the third question was Top, but I'll ask her again. Maybe I just didn't hear it correctly. –  KarenVO Apr 28 '12 at 9:59
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