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Under what conditions on the topological space $X$ is the overcategory $\mathbf{Top}/X$ of topological spaces over $X$ equivalent to a full subcategory of $\mathbf{Top}$? Surely if $X$ terminal i.e. a point, but is that the only case?

Obviously I would be happiest with a general criterion valid for some large clas of categories, telling us when $\mathbf C/X$ embedds fully faithfully into $\mathbf C,$ but really it is the case $\mathbf C=\mathbf {Top}$ that I am interested in.

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    $\begingroup$ That seems like an unnatural thing to ask for in the absence of a natural choice of such an embedding. Why do you want to know? $\endgroup$ Commented Nov 21, 2014 at 4:45
  • $\begingroup$ Because any canonical functor from an overcategory into a category that I can think of is not an embedding, so the question came to me if there can be some very contrived unnatural way to set such an embedding up. $\endgroup$ Commented Nov 21, 2014 at 5:08
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    $\begingroup$ But... why would you want such a thing? $\endgroup$ Commented Nov 21, 2014 at 5:09
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    $\begingroup$ I was doing something with overcategories of topological spaces and at one point it occured to me that it would be fabulous if I could just embedd them into topological spaces. "But clearly that can't be done" was my first thought, but then I could not find a reason why, so I posted it here. A stupid little musing, nothing very deep. $\endgroup$ Commented Nov 21, 2014 at 5:15

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This is true only if $X$ has at most one point. Suppose $i:\mathbf{Top}/X\to \mathbf{Top}$ is a full embedding. Write $Id$ for the terminal object of $\mathbf{Top}/X$, the identity map $X\to X$. Then $i(Id)$ must have only one continuous self-map, and hence has at most one point (since any constant map is always continuous). If $i(Id)$ is empty it follows easily that $X$ must be empty, so we may assume $i(Id)$ has one point. Now let $x,y\in X$ be any two points. Let $A$ and $B$ be the inclusion maps of $\{x\}$ and $\{y\}$ into $X$, respectively, considered as objects of $\mathbf{Top}/X$. There are no maps $Id\to A$ or $Id\to B$, so since $i$ is full, $i(A)$ and $i(B)$ must both be empty. Since $i$ is full, this means $A$ and $B$ must be isomorphic, so we must have $x=y$. Thus $X$ can have at most one point.

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