# Morphisms that are injective and surjective but not isomorphisms

I'm reading Silverman's Arithmetic of Elliptic Curves and read that isogenies are group homomorphisms and if $$\phi:E_1\longrightarrow E_2$$ is a non-zero isogeny between elliptic curves,its kernel $$Ker\phi=\phi^{-1}(O)$$ where $$O$$ is the point at infinty of $$E_2$$.

Now suppose that $$\phi\in End(E)$$ be a non-zero isogeny for some elliptic curve $$E$$.Then,applying the first isomorphism theorem we have, $$E/Ker\phi\cong \phi(E)$$ Since every isogeny is a morphism between smooth curves and $$\phi\neq O$$ it is surjective,which implies,$$Ker\phi=O$$.That means $$\phi$$ is injective.But $$End(E)-{O}\neq Aut(E)$$,so $$\phi$$ need not be an isomorphism.

1.) Can someone give me an example of some isogeny and elliptic curve where I get to know how it fails to have an inverse,i.e, not be an isomorphism?

Edit 1:
2.) I have another question regarding the argument above.Silverman also defines a multiplication by m-map as follows:

Let $$m\in\mathbb{Z}$$.Define the map $$[m]:E\longrightarrow E, \ P\rightarrow P+...+P$$ m times if m>0 and $$(-P)+...+(-P)$$, $$-m$$ times if m<0 and $$[0]P=O$$.Then $$[m]$$ is a non-constant isogeny for $$m\neq0$$.

Now this means $$[m]\in End(E)$$ and by the above argument it implies $$Ker[m]=E[m]={O},\forall m\in\mathbb{Z}-\{0\}$$ which is not true.So what is actually going wrong in my argument.

Edit 2: Ok,so the argument $$E/Ker\phi\cong E \implies Ker\phi={O}$$ is not true since $$E$$ may be an infinite group and for infinite groups this maybe the case:Does $G\cong G/H$ imply that $H$ is trivial?

• Why do you think that the map being surjective implies that $\ker \phi = 0$? We have $E/\ker \phi \cong E$, but this is still very much possible for this to happen even when $\ker \phi \neq 0$. Jul 13, 2023 at 0:47
• Yes, I have mentioned the mistake in my most recent edit(edit 2.). Jul 13, 2023 at 6:30