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I have proved that my category $\mathbf{Fcd}$ has small products. (Correction, it seems it has no terminal object that is empty product. I have overlooked this earlier in my proof.) No, it indeed has a terminal object.

Next I tried to prove that it has an exponential object (and so it cartesian closed).

I first conjectured that the exponential object $Y^X = \operatorname{id}^{\mathbf{Fcd}}_{(Y^X)}$ (where $Y^X$ in the right part of the equation means the set of graphs of functions in $Y^X$).

But if it is an exponential object, then there are isomorphisms between $Z \rightarrow \operatorname{id}^{\mathbf{Fcd}}_{( Y^X)}$ and $Z \times X \rightarrow Y$, which seems not a case.

So now I suspect that the category $\mathbf{Fcd}$ has no exponential object.

As such, how to prove that there are no exponential object in a category?

You may read about my category $\mathbf{Fcd}$ at http://www.mathematics21.org/binaries/product.pdf (to fully understand it, you need first read my book: http://www.mathematics21.org/algebraic-general-topology.html). Well, in order to be able to answer my question, you may probably don't need to read my writings first. I post links to my writings for reference.

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    $\begingroup$ Proof by contradiction usually works... $\endgroup$ – David Roberts Sep 8 '13 at 19:10
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    $\begingroup$ If your category has some colimits (such as an initial object, coproducts, and coequalizers), you might try to determine whether the functor $X \times -$ (for any object $X$) preserves them. This is the classic test for necessity; if $X \times -$ fails to preserve some such colimit, it cannot be cartesian closed. $\endgroup$ – Todd Trimble Sep 8 '13 at 19:13
  • $\begingroup$ @ToddTrimble: Thanks, I will follow your advice to investigate colimits $\endgroup$ – porton Sep 8 '13 at 19:14
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    $\begingroup$ @ViditNanda I think that's an English grammar thing. It would be silly to try for absolutely no exponentials, since the terminal object is always exponentiable. $\endgroup$ – Todd Trimble Sep 8 '13 at 19:42
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    $\begingroup$ @ViditNanda Well, he did say it has all small products. I've not studied the category of funcoids myself. $\endgroup$ – Todd Trimble Sep 8 '13 at 20:22
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As a first step, it may help to choose some relatively simple examples of $X$ and $Y$. If you can find a characterization of objects in your category as representable presheaves among all presheaves, or even some easy-to-understand necessary conditions, it may help to exclude the representability of the presheaf $Y^X$.

Here's an example from algebraic geometry. In the category of affine varieties over the complex numbers, set $X$ and $Y$ to be the affine line $\mathbb{A}^1 = \operatorname{Spec} \mathbb{C}[x]$. Then for any finitely generated integral domain $R$ over $\mathbb{C}$, the maps from $\operatorname{Spec} R$ to $Y^X$ are identified with maps from $\operatorname{Spec} R \times \mathbb{A}^1$ to $\mathbb{A}^1$, or equivalently $\mathbb{C}$-algebra maps $\mathbb{C}[x] \to R[x]$, which are determined uniquely by the image of $x$. The presheaf $Y^X$ is then the colimit of an increasing system of presheaves $\operatorname{Spec} R \mapsto \bigoplus_{i=0}^n Rx^n$ represented by finite dimensional affine spaces. It isn't representable, because affine varieties satisfy the condition that they only admit injective presheaf maps from affine spaces of bounded dimension.

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    $\begingroup$ The functor $F:R \rightsquigarrow R[x]$ on $k$-algebras ($k$ a ring) is not representable on the category of all $k$-schemes as well. Say $M$ represents $F$ and choose $m \in M$. On an open affine neighborhood $U$ of $m$ the "universal polynomial" has degree below some $n$, so $U \hookrightarrow M$ factors through the monic $j_n:\mathbf{A}^n \rightarrow M$. Factoring $j_n$ through $j_{n+1}$, the closed immersion $\mathbf{A}^n \hookrightarrow \mathbf{A}^{n+1}$ ("first $n$ coordinates") would have non-empty interior, absurd. Using etale maps, likewise there's no $M$ as an algebraic space. $\endgroup$ – Marguax Sep 8 '13 at 20:32

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