According to ESCARDÓ-LAWSON-SIMPSON paper 'Comparing cartesian closed categories of (core) compactly generated spaces' The following four propositions are true:

  • A topological space $X$ is exponentiable iff $X$ is core-compact.

  • The category of core-compact topological spaces is not cartesian closed, because even though when $X$ and $Y$ are core-compact, we can build the exponential object $Y^X$, but $Y^X$ will not necessarily be itself core-compact.

  • The category of Core-compactly generated spaces is cartesian closed.

  • The category of Core-compactly generated spaces contains the category of core-compact spaces.

It seems that there is a contradiction if the four propositions are all true. A sub-category of the category of topological spaces cannot be cartesian closed if it is not strictly included in the sub-category of core-compact spaces. What Am i getting wrong here ?

| cite | improve this question | | | | |
  • $\begingroup$ The way I read this is that an exponential $Y^X$ of two core compact spaces may fail to be core-compactly generated. $\endgroup$ – Andrej Bauer Dec 5 '11 at 14:10
  • 4
    $\begingroup$ "Exponentiable" is relative to the ambient category, and a non-exponentiable object in a category $C$ may turn out to be exponentiable in some subcategory. $\endgroup$ – Qiaochu Yuan Dec 5 '11 at 14:11

Andrej and Qiaochu are right. Let me elaborate. If X and Y are core-compactly generated (ccg for short), then the exponential in the category of topological spaces need not exist (it exists iff X is core compact). But the exponential always exists in the category CCG. You have to take into account the facts that (1) the finite products change when you pass to a subcategory (there are fewer spaces to test the universal property), (2) consequently the exponentials change too. Thus CCG is cartesian closed without contradicting the fact that the exponential of two ccg spaces in CCG is not an exponential in the whole category of topological spaces.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.