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Is it possible to construct a category such that there is a monomorphism and epimorphism that is neither injective nor surjective(set-theoretically).

I notice that there is a solution in which a monomorphism and epimorphism is not surjective. I wonder if a morphism always has to be "injective" as long as it is a monomorphism and epimorphism.

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    There are categories whose morphisms are not even maps; what would it mean for one of those to be "neither injective nor surjective"? – 2017-01-23
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    @HenningMakholm I want a map which is neither surjective nor injective. Of course, the category is not "Set", in which the statement is true. – 2017-01-23
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    @TiWen The last sentence of your question suggests that you believe the morphisms of a category are set-theoretic functions maybe satisfying some properties or with extra structure. Henning is saying that this isn't the case. For example, given a partial order $\leq$, you can make a category where $\text{Hom}(X,Y) = \{(X,Y)\}$ if $X \leq Y$ and $\{\}$ otherwise. A morphism in this category is just a pair of objects. What does it mean for a pair of objects to be "injective"? (Note, in this example all morphisms are mono and epi.) – 2017-01-23

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Consider this silly category:

Object: Sets $A$ such that $\{0,1\}\subseteq A\subseteq \mathbb N$.

Morphisms: Maps $f:A\to B$ with the properties $$ f(0)=f(1)=0 \\ x\notin\{0,1\}\Rightarrow f(x)\notin\{0,1\} $$

Composition: Function composition.

Identity: $\mathrm{id}(x) = \begin{cases} 0 &\text{if }x=1 \\ x & \text{otherwise} \end{cases}$

Then no morphism at all is either injective or surjective as a map -- but there are plenty of epimorphisms and monomorphisms, such as all the identities.

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Yes, for instance, define a subcategory of $\mathbf{Set}$ whose only objects are $$A=\{1,2\}\qquad\text{and}\qquad B=\{3,4\}$$ and morphisms are given as $$\begin{align}\operatorname{Hom}(A,A)&=\{\operatorname{id}_A\},\\\operatorname{Hom}(A,B)&=\{f\},\\\operatorname{Hom}(B,A)&=\emptyset,\\\operatorname{Hom}(B,B)&=\{\operatorname{id}_B\},\end{align}$$ where $f:A\to B$ is defined by $f(1)=f(2)=3$. Then $f$ is a monomorphism and an epimorphism, but it is neither injective nor surjective.