Let $E:y^2=x^3+x$ be an elliptic curve over $\Bbb{Q}$.
Let $E':y^2=x^3-4x$ be an another elliptic curve.
There is an isogeny $\phi : E\to E'$ given by $(x,y)\to (\frac{y^2}{x^2},\frac{y(1-x^2)}{x^2})$.
This is called $2$-isogeny, because its degree is exactly $2$. Thus $Ker\phi\cong \Bbb{Z}/2\Bbb{Z}$.
Also, $\phi$ is surjective because it is nonzero morphism between curves.
But form first isomorphism theorem, $E(\Bbb{Q})/ker\phi \cong E'(\Bbb{Q})$・・・①.
But according to LMFDB, $E(\Bbb{Q})\cong E'(\Bbb{Q})\cong \Bbb{Z}/2\Bbb{Z}$.
Thus from ①, $0 \cong \Bbb{Z}/2\Bbb{Z}$. This is unreasonable.
Where did I go wrong ?
The existence of an isogeny $\phi:E\to E'$ only implies the existence of a (surjective) homomorphism between the groups of points over an algebraic closure: $$E(\overline{\Bbb{Q}})/\mathrm{Ker}(\phi)\simeq E'(\overline{\Bbb{Q}}).$$ It also maps rational points to rational points, but that mapping need not be surjective.