Special linear group is fully characteristic in general linear group

Statement
Let $$k$$ be a field and $$n$$ be a natural number. Let $$GL_n(k)$$ denote the group of invertible $$n \times n$$ matrices over $$k$$ under multiplication and $$SL_n(k)$$ denote the subgroup comprising matrices of determinant one. $$SL_n(k)$$ is a fully characteristic subgroup of $$GL_n(k)$$: every endomorphism of $$GL_n(k)$$ restricts to an endomorphism of $$SL_n(k)$$.

Facts used

 * 1) uses::Commutator subgroup of general linear group is special linear group: This is true except in the case where $$n = 2$$ and the field $$k$$ has exactly two elements.
 * 2) uses::Commutator subgroup is fully characteristic

For the field with two elements
In this case, every element of $$GL_n(k)$$ has determinant one, so $$SL_n(k) = GL_n(k)$$. Since every group is fully characteristic as a subgroup of itself, $$SL_n(k)$$ is a fully characteristic subgroup of $$GL_n(k)$$.

For other cases
In all other cases, the result follows from facts (1) and (2).