# Why is the kernel of reduction contained in the image of this isogeny? (From a paper of Cassels)

$$\newcommand{\fp}{\mathfrak{p}} \renewcommand{\phi}{\varphi}$$ I'm currently trying to understand a paper of Cassels [see page 189 for the relevant content, esp. (4.9)] and I've hit a little snag. The set up is as follows (although I've modernized notation and slightly changed the context):

Let $$K$$ be a number field, $$\fp$$ a (finite) prime of $$K$$, and $$k_\fp$$ the residue field. Fix an algebraic (resp. separable) closure $$\bar{K}_\fp$$ (resp. $$\bar{k}_\fp$$) of $$K_\fp$$ (resp. $$k_\fp$$). Let $$E,E'$$ be elliptic curves defined over $$K_\fp$$ with good reduction at $$\fp$$, and suppose we have an isogeny $$\phi:E\to E'$$ such that $$\fp\nmid\deg\phi$$. Let $$\tilde{E},\tilde{E}'$$ be the reduced curves over $$k_\fp$$. Then by some standard result (I think it's in Silverman somewhere) there is a separable isogeny $$\tilde{\phi}:\tilde{E}\to \tilde{E}'$$ which is the 'reduction' of $$\phi$$—that is, the obvious square commutes and the $$\bar{k}_\fp$$-points of the kernel of $$\tilde{\phi}$$ are the reduced $$\bar{K}_\fp$$-points of the kernel of $$\phi$$.

Now for the bit I don't get. According to Cassels, "because $$\tilde{\phi}$$ is separable, we have $$E'_1(K_\fp)\subseteq\phi(E(K_\fp))",$$ where $$E'_1$$ is the kernel of reduction $$E'\to\tilde{E}'$$. How does this follow?