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EDIT: The original question appears to have been worded much too abstractly in terms of constant and coconstant morphisms to admit a precise answer (for details see history), the new question has been restated in a much more concrete form after receiving comments.

Restated question: Can morphisms in a concrete category admit mappings that are not functions?

It arose when dualizing a constant function $f:X\longrightarrow Y$ in $\mathcal C$ into a dual morphism $F(f):F(Y)\longrightarrow F(X)$ in $\mathcal B^{op}$ under a contravariant dually equivalent functor $F:\mathcal C\longrightarrow \mathcal B^{op}$ with $\mathcal C \simeq \mathcal B^{op}$ (dually equivalent). $F(f)$ then should map a constant $c\in Y$ in such a way that $F(f)(c):F(Y)\longrightarrow F(X)$ outputs its entire codomain $F(X)$, which is at the same time both the image and the output of $F(f)(c)$. This appears to be a multifunction rather than a function.

For more concreteness, $\mathcal B^{op}$ can be taken as $Set^{op}$. A construction is given on Wikipedia on how to embed $Set^{op}$ into the category Rel of sets and relations. Both Rel and $Set^{op}$ are known to be concrete categories. Then the morphism $F(f)$ in $Set^{op}$ that is dual to a constant function $f$ in $\mathcal C$ can be a multifunction, because relations need not be functions. Some more specifics and references are given in my comments to this post.

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    $\begingroup$ You don't define anything here. You just write down $F(c)$ and the property of $F(c)$ which follows from the property of $c$. And that's correct. One calls this a coconstant morphism. (Why not nstant ...) Google gives plenty of material. $\endgroup$ – HeinrichD Apr 19 '17 at 18:24
  • $\begingroup$ Thanks @HeinrichD. I left a follow up question in the original post. $\endgroup$ – Dawid K Apr 21 '17 at 2:25
  • $\begingroup$ How do you define an element of $F(C)$? Remember the definition of a category, the objects are no sets. Maybe you want to talk about generalized elements. But in any case, one generalized element is mapped exactly to one generalized element. This is what maps do. $\endgroup$ – HeinrichD Apr 21 '17 at 6:41
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    $\begingroup$ In the sense of "global points", in the sense of morphisms $1\to X$, then it will be the case that $f:X\to Y$ will, by composition simply take that point to another point, and coconstant morphisms don't rule out that the resulting points may be distinct. (And if such a point were epimorphic, it would be an isomorphism.) For a concrete category, it should be obvious that the idea of mapping an element to every element in the codomain is nonsensical: the underlying morphisms of a concrete category are functions, and cannot map that way. $\endgroup$ – Malice Vidrine Apr 22 '17 at 15:33
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    $\begingroup$ A) If you come up with a new question, you should probably post a new question, not just edit an older post. B) The former version of $\mathbf{Rel}$ you describe is the more standard one; the latter I've never seen outside Abstract and Concrete Categories. $\endgroup$ – Malice Vidrine Apr 23 '17 at 21:13
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To the most recent version of the question, the answer is that it depends on what you mean by "concrete category". The most common usage of the term means, specifically, a category equipped with a faithful functor to $\mathbf{Set}$, so the answer would be no. If you've been reading Abstract and Concrete categories, what most folks now call "concrete categories" are what those authors refer to as "constructs."

There is also the more general notion of "concrete over $\mathcal D$," meaning it's a category equipped with a faithful functor into $\mathcal{D}$. In this case, sure, you could have a concrete category over $\mathbf{Rel}$, or over the trivial category (as with preorders), or whatever else. For instance, $\mathbf{Set}$ is concrete over $\mathbf{Rel}$ in this sense, or over $\mathbf{Eff}$.

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  • $\begingroup$ This perhaps nails it. I must have been thinking about concrete over $Set^{op}$ or something similar. Maybe there is something funny about embedding $Set^{op}$ into $Rel$. Thanks for the input. $\endgroup$ – Dawid K Apr 23 '17 at 21:35

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