For an isogeny of abelian varieties $f : X \to Y$, is $Y = X/ \operatorname{ker}f$? Let $k$ be a field, $f : X \to Y$ be an isogeny of $k$-abelian varieties.
Then there exists the canonical separable isogeny $\pi : X \to X/\operatorname{ker}f$, such that $X/\operatorname{ker}f$ is a $k$-abelian variety, and on $\overline{k}$-valued points, $\pi$ is the natural quotient map of groups with the kernel $\operatorname{ker}f$.
By III.4.1 of Silverman's The Arithmetic of Elliptic Curves, there exists an isogeny $g : X/\operatorname{ker}f \to Y$ such that $f = g \circ \pi$.
Now, is $g$ an isomorphism (of varieties)?
The Corollary 1 of section 12 of Mumford's Abelian Varieties says this is true.
However, if so, I think that every isogeny becomes separable: $\pi$ is separable, and $g$ is an isomorphism, in particular separable, thus $f = g \circ \pi$ is also separable.
 A: I confirm the part corresponding to Silverman's book: in there $\operatorname{Ker}(f)$ for $f: X\to Y$ an isogeny of elliptic curves means the set of $\bar K$-points of $X$ that go to $0\in Y$. This is a finite (étale) subgroup scheme of $X$, and there exists a unique separable isogeny $g: X\to X'$ such that $\operatorname{Ker}(f)=\operatorname{Ker}(g)$ (by proposition III.4.12 in Silverman's). Moreover, by corollary 4.11, there exists a unique isogeny $h:X'\to X$ such that $f=h\circ g$. This isogeny $h$ is purely inseparable, and $\operatorname{Ker}(h)=\{0\}$, but $h$ is not aan isomorphism. 
On the other hand, in books with scheme theoretic flavour (such as Mumford's book), $\operatorname{Ker}(f)$ means the kernel group-scheme, which is a finite subgroup scheme of $X$. One has that $\operatorname{Ker}(f)(\bar K)$ is what Silverman called $\operatorname{Ker}(f)$ above. 
For example, if $K$ has characteristic $p$, and $f=[p]$ is multiplication by $p$, then $\operatorname{Ker}(f)(\bar K)$ has either $p$ points (if $E$ is ordinary) or $1$ point (if it is supersingular). In the first case $g:X\to X'$ has degree $p$, and also $h$. In the second case $g=id$ is the identity, and $h=f$. 
By the way, the fact that $f=h\circ g$, with $g$ separable and $h$ purely inseparable is a version of the result that says that any finite field extension $L'/L$, there exists an intermediate extension $L'/L_s/L$ with $L_s/L$ separable and $L'/L_s$ purely inseparable: $L_s$ is the set of separable elements in $L'$. 
