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The converse holds so if $G_1$ is cyclic it is shown that also $G_2$ is cyclic. I can not take the inverse of the function $g$ since the function is not injective ( i wanted to show that the generator of $G_2$ is mapped to the generator of $G_1$). Since the identity element of $G_2$ is mapped to $Kerg$ , then the generator of $G_2$ is mapped to one of the partitioned classes of $G_1$ and because of structure perserverance $G_1/Kerg$ isomorphic to $G_2$ and hence is cyclic. From this point on I am stuck since I can not prove that if $G_1/Kerg$ is cyclic then $G$ is cyclic?
Thank you,

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    Do you mean homomorphism? Epimorphism? Endomorphism is the wrong word here. – 2017-02-09
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    If all you mean is homomorphism, then no. Just map all of $G_1$ to the identity. – 2017-02-09
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    I thought it might mean "injection", because at least that case would be interesting; homomorphism is bad as lulu points out; epimorphism is bad by mapping $S_3$ to $\Bbb Z / 2 \Bbb Z$. For an injection, I don't see quite such an obvious counterexample anywhere, but maybe I just need more caffeine. – 2017-02-09
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    If $f:G \to H$ is an injection, then $f(G)$ is a subgroup of $H$ and isomorphic to $G$. Since $H$ is cyclic, all subgroups are cyclic, and so $f(G) \cong G$ is cyclic. – 2017-02-09
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    Thank you for the replies , I meant epimirphism instead. I am not a native English speaker. – 2017-02-09
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    Thanks, @DKS: enlightenment is even better than a cup of coffee. :) – 2017-02-09
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    Happy to assist! – 2017-02-09

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This is false. Map $S_n \to \mathbb{Z}_2$ ($n\geq 3$) using the sign of the permutation. The target is cyclic (and in fact, $\mathbb{Z}_2$ is isomorphic to a subgroup of $S_n$ for $n\geq 2$, just take $\langle (1,2) \rangle$ as the subgroup - just in case you need an endomorphism specifically), but the source is decidedly not.

To put in an edit quickly, if $f$ is an injection, then this is true. See my comment.

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    What do you mean by endomorphism? If you have an endomorphism and either domain or codomain is cyclic then the other is too by definition. – 2017-02-09
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    An endomorphism is a homomorphism from the algebraic object to itself. It need not be injective. The wording of the question matters, as the image could be cyclic. – 2017-02-09
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    How is that a counter example? The domain and codomain are still the same, hence cyclic simultaneously. – 2017-02-09
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    I misinterpreted you and edited my comment. It's due to ambiguity in the question. When the asker said endomorphism from $G_1$ to $G_2$, I assumed $G_2 < G_1$. So the image could be cyclic, while the total target space could be otherwise. – 2017-02-09
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If by epimorphism of groups one understands a surjective homomorphism the the existence is sometimes possible, sometimes not. A counterexample has already been given in another answer. An example where it exists is given by $G_1 = C_{2m}$, a cyclic group of order $2m$ and $G_2 = C_m$, a subgroup of $G_1$, the epimorphsm given by the projection $g \mapsto g^2$.