i was reading my notes and it states that the kernel of a function allows you to check if the function is injective.
how do you check with a kernel? isint elements in the kernel, elements that through the function get mapped to the identity?
im studying group homomorphisms now.
in this case, if it is a function from A to B, where the identity of B =0, then the element in A has to be 0 since u=0 (from your above statement) right?
what if the element in A that get mapped to 0 in B is not 0, wont u be not equals to 0? in such case, Tu=0 but u wont be zero?
thanks!
My above proof can be quite easily changed to look at group homomorphisms (just change a+b to ab, and a-b to ). So I'll use group notation.
Is your question basically,
Is it possible that ?
The answer is `no'.
Let . Then . Thus, ( is an idempotent). Then, . Thus, for a group homomorphism.
The same holds for linear maps, etc.