The entry lists some of the common examples of epis (epimorphisms). The examples also demonstrate some of the techniques used in finding epis.

1. In Set, the category of sets, the epis are exactly the onto functions. First, suppose $f:A\to B$ is onto and that $g,h:B\to C$ are functions such that $g\circ f=h\circ f$ . Then for any $b\in B$ , there is $a\in A$ such that $f(a)=b$ since $f$ is onto. This means that $g(b)=g(f(a))= h(f(a))=h(b)$ , or $g=h$ , showing that $f$ is epi. Conversely, suppose $f:A\to B$ is epi. Define functions $g,h:B\to \lbrace 0,1\rbrace$ as follows: $g(x)=0$ for all $x\in B$ , and $h(x)=0$ if $x\in f(A)$ and $h(x)=1$ otherwise. Then $g(f(a))=0=h(f(a))$ . This means that $g=h$ , or $x\in f(A)$ for all $x\in B$ . In other words, $f$ is onto.
2. In Ab, the category of abelian groups, the epis are exactly the onto abelian group homomorphisms. If $f$ is onto and $g\circ f =h\circ f$ , then for any $b\in B$ , there is $a\in A$ such that $f(a)=b$ . This means that $g(b)=g(f(a))=h(f(a))=h(b)$ , or $g=h$ , showing that $f$ is epi. On the other hand, suppose $f:A\to B$ is epi. Define $g,h:B\to B/f(A)$ as follows: $g(x)=f(A)$ and $h(x)=x+f(A)$ for all $x\in B$ . Then $g(f(a))=f(A)=f(a)+f(A)=h(f(a))$ . This implies that $g=h$ , so that $x\in f(A)$ for all $x\in B$ , or $f$ is onto.
3. In Top, the category of topological spaces, the epis are exactly the surjective continuous functions. If $f$ is onto and $g\circ f=h\circ f$ , then then for any $b\in B$ , there is $a\in A$ such that $f(a)=b$ . This means that $g(b)=g(f(a))=h(f(a))=h(b)$ , or $g=h$ , showing that $f$ is epi. On the other hand, suppose $f:X\to Y$ is epi. Equip $\lbrace 0,1\rbrace$ with the trivial topology. Define $g,h:Y\to \lbrace 0,1\rbrace$ as in example 1 above. Then $g$ and $h$ are both continuous. We also have $g(f(a))=0=h(f(a))$ , so that $g=h$ , or $x\in f(A)$ for all $x\in B$ . Therefore, $f$ is onto.

Not all epimorphisms are surjections. For example, in the category CommRng of commutative rings with 1, the natural injection $i:\mathbb{Z}\to \mathbb{Q}$ is clearly not a surjection, and yet it is epimorphic. To see this, let $R$ be any commutative ring with characteristic $0$ . Suppose $g,h:\mathbb{Q}\to R$ are ring homomorphisms such that $g\circ i=h\circ i$ , in other words, $g(n)=h(n)$ for all $n\in \mathbb{Z}$ . Set $f:=g-h$ . Then $f(n)=0$ for all $n\in \mathbb{Z}$ . Then $0=f(n)=mf(n/m)$ , where $m$ is an arbitrary positive integer. Since $\operatorname{char}(R)=0$ , this shows that $f(n/m)=0$ . Since $n/m$ is an arbitrary rational number, $f=0$ , or $g=h$ . Hence $i$ is an epi.

For another counterexample, it can be shown that in HausTop, the category of Hausdorff topological spaces and continuous functions, the epimorphisms are precisely the continuous functions with dense images. As such, surjections are not a requirement.