Isomorphism and $\mathrm{id}$ In a category I have two objects $a$ and $b$ and a morphism $m$ from $a$ to $b$ and one $n$ from $b$ to $a$. Is this always an isomorphism? Why is it emphasized that this has to be true, too: $m \circ n = \mathrm{id}_b$ and $n \circ m = \mathrm{id}_a$?

I am looking for an example in which the id-part is not true and therefore $m$ and $n$ are not isomorphic.
If you have no restrictions on m and n, then clearly they cannot be isomorphisms in general.  For instance, take any two groups G and H and let m: G --> H, n: H --> G be the zero homomorphisms.

Even if you say that n and m are monomorphisms, then it is still not true in general that they are isomorphisms.  I believe it is true however if your category is one whose objects are sets with additional structure. See this question: If $A$ is a subobject of $B$, and $B$ a subobject of $A$, are they isomorphic?.
To make this perhaps even clearer, say we have the objects $a=\{x, y\}$ and $b=\{z\}$.  Define $m\colon a\to b$ by $x\mapsto z$, $y\mapsto z$ (the only possible morphism here).  Define $n\colon b\to a$ by $z\mapsto x$.  But, of course, $n\circ m$ doesn't equal $\mathrm{id}_a$, since $(n\circ m)(y)=n(m(y))=n(z)=x$, $(n\circ m)(x)=n(m(x))=n(z)=x$, so we have $n\circ m\colon a\to a$ is $x\mapsto x$, $y\mapsto x$.  

Perhaps if the cardinality of objects $a$ and $b$ equal each other, then a morphism $m$ from $a$ to $b$ and a morphism $n$ from $b$ to $a$, then both $m$ qualifies as an isomorphism and $n$ qualifies as an isomorphism also.