Are the values of generators of a homomorphism of groups distinct? Given the fact that a homomorphism $ f:G \rightarrow F $  is determined by ist values at a set of generators $E$ of $G$, $G=EE^{-1}$, it means that $f(x_{i})=b_{i}, x_{i}\in E$ and $b_{i} \in F$. 


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*Since a homomorphism must not be an isomorphism, does this means that the $b_{i}$ must not all be distinct ? Say that $f(x_{1})=f(x_{2})$, for $x_{1} \neq x_{2} $, two generators of $G$. It follows that $x_{1}, x_{2} \in ker f$. I am not quite sure what consequences this induces in the group $F$. 

*In the case of the homomorphism $Z\rightarrow G$, it is characterised by ist value on the generator $1$ of $Z$. How many endomorphisms and isomorphisms $f: Z\rightarrow Z$ are there ? If i set $f(1)=1 $ it is clear that $f$ will be an isomorphism since $f(n)=nf(1).$ 
Can somebody help me out to answer these questions. Thanks for any comment.
 A: 1) A group homomorphism can be an isomorphism.  In fact an isomorphism is by definition a homomorphism that also happens to be a bijection.  There are homomorphisms that are not isomorphisms, but all isomorphisms are homomorphisms.  Isomorphisms are a special kind of homomorphism.
So no, it is not required that two basis elements get sent to the same place.  Also, by the way, if you have a homomorphism that is not an isomorphism it is still not required that that happens.  For example $\mathbb Z^2$ is generated by $(1, 0)$ and $(0, 1)$.  The map $\mathbb Z^2 \to \mathbb Z$ defined by $(a, b) \mapsto 3a - 2b$ sends the generators to $3$ and $-2$ and these are distinct but the map is still not an isomorphism.  For example $(2, 3)$ is in the kernel.
2) There are countably many endomorphisms of $\mathbb Z$, one for each element of $\mathbb Z$.  The group $\mathbb Z$ is free of rank $1$ (we can take the number $1$ to be the generator).  This means that to specify a homomorphism $\mathbb Z \to G$ you just have to specify an element $g \in G$ to be the image of $1$ and any choice of $g$ gives you a well defined homomorphism.  So there are $|G|$ many homomorphisms $\mathbb Z \to G$.  A homomorphism $\mathbb Z \to \mathbb Z$ will only be an isomorphism if $1$ is sent to a generator of $\mathbb Z$.  There are only two generators of $\mathbb Z$, they are $1$ and $-1$ so there are only two isomorphisms $\mathbb Z \to \mathbb Z$.  The identity $n \mapsto n$ and negation $n \mapsto -n$.
