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$ \newcommand{\cat}{\mathbf} \newcommand{\RR}{\mathrm R} $

If there exist a morphism $$f:\RR \rightarrow [0,+\infty] ,f : x\mapsto x^2+1$$ in a category $\cat{Set}$. In particular, $f$ is not injective nor a surjective.

In the category of $\cat{Set^{op}}$, the $f$ change into $$f^{op}:[0,+\infty]\rightarrow \RR $$

This raises a question : $$f^{op}:x\mapsto ??? $$

The definition must satisfy that: $$ \cat {(C^{op})^{op}} = \cat C $$

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    The discussion [at this other question](http://math.stackexchange.com/questions/2092872/do-opposite-categories-always-exist) might be helpful to you. Short version: morphisms (and objects) are in general *purely abstract*; even though one category might have a very "concrete" interpretatoin, its opposite (which is defined purely formally) need not. – 2017-02-11

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The morphism $f^{\mathrm{op}}$ is still the same function as $f$. In a category, the morphisms can be anything; they don't need to be functions. (For example, think about treating a poset like a category: there, the morphisms aren't functions either.) In particular, it is perfectly allowed to make up a category $\mathsf C$, where the objects of $\mathsf C$ are sets, and a morphism $f: X \to Y$ is actually a function $f: Y \to X$. (Can you see what composition has to be?) Then, the category $\mathsf C$ is just $\mathsf{Set}^{\mathrm{op}}$.

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    Why the morphism $f^{op}$ is still the same function as $f$? they have diffrent domain and codomain. – 2017-02-10
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    @xiang A morphism in $\mathbf{Set}^{\mathrm{op}}$ is *not* a function. – 2017-02-10
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    $$f^{op}:x\mapsto \begin{cases} \pm \sqrt{x-1},&x>1 \\ nothing,&0\leq x<1 \end{cases}$$. Is it right? – 2017-02-10
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    That answer is mistaken. A morphism need not be a function, because an object in a category need not be a set. A morphism $f$ in a category is indeed the same as $f^{op}$ in the dual category, but its domain and codomain are switched. – 2017-02-10
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    @xiang: The domain of $f^\text{op}$ as a function need not be the same thing as the domain of $f^\text{op}$ as a morphism of a category. – 2017-02-11
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    @SixWingedSeraph My answer is not function, because in function each x maps to only one value, but in $f^{op}$,it x can maps to 0,1,2…… or more values . – 2017-02-11
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    @xiang: My point is that the morphism $f$ may not be a function. It may merely be an arrow drawn on a piece of paper. There are examples in A slow introduction to categories (http://www.abstractmath.org/Word%20Press/?p=10130) and in the Wikibooks article on categories. There is no question that you can assume that $f$ maps elements to elements -- its domain may not have elements. Similarly, $f^{op}$ may not "map" anything to anything, multivalued or otherwise. – 2017-02-12
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In $\mathbf{Set}^\text{op}$, $\hom(X, Y)$ is the set of all functions from $Y$ to $X$.

That said, there is an equivalent description: that $\hom(X,Y)$ is the set of all functions $\mathcal{P}(X) \to \mathcal{P}(Y)$ that preserve unions and intersections.

More precisely, $\mathbf{Set}^\text{op}$ is equivalent to the category whose objects are complete atomic boolean algebras and whose morphisms are functions that preserve all unions and intersections.

The bijection between the two descriptions associates to a function $f : Y \to X$ the function $f^* : \mathcal{P}(X) \to \mathcal{P}(Y)$ defined by $$f^*(S) = \{ y \in Y \mid f(y) \in S \} $$

It may be interesting to note that there are naturally occurring categories where morphisms involve functions going the "wrong way". For example, in Top, the category of topological spaces and continuous maps, the homomorphisms $(|X|, \mathcal{S}) \to (|Y|, \mathcal{T})$ involve functions $|X| \to |Y|$ (i.e. a mapping on points) and also functions $\mathcal{T} \to \mathcal{S}$ (the inverse image of an open set is open).

A related notion is that of a locale. It's like a topological space, but without the set of points: a locale is just the frame of open 'sets'. So, in the category Loc of locales, a locale morphism $\mathcal{S} \to \mathcal{T}$ is a (structure preserving) function $\mathcal{T} \to \mathcal{S}$!