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?
What are we "in" here? Linear maps? Group homomorphisms?
The proof is pretty much the same in every context through.
Let $\displaystyle T$ be a linear map.
Then $\displaystyle T$ is injective
$\displaystyle \Leftrightarrow (Tx = Ty \Rightarrow x=y) $
$\displaystyle \Leftrightarrow (Tx-Ty = 0 \Rightarrow x=y) $
$\displaystyle \Leftrightarrow (T(x-y) = 0 \Rightarrow x = y) $
$\displaystyle \Leftrightarrow (T(x-y) = 0 \Rightarrow x-y = 0) $
$\displaystyle \Leftrightarrow (Tu = 0 \Rightarrow u =0) $
$\displaystyle \Leftrightarrow kerT = \{0\}$.
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 $\displaystyle ab^{-1}$). So I'll use group notation.
Is your question basically,
Is it possible that $\displaystyle 1 \not\in ket(\phi)$?
The answer is `no'.
Let $\displaystyle 1 \phi = a$. Then $\displaystyle a^2 = (1\phi)^2 = (1^2)\phi = 1\phi = a$. Thus, $\displaystyle a^2=a$ ($\displaystyle a$ is an idempotent). Then, $\displaystyle a^2 = a \Rightarrow a^2a^{-1} = aa^{-1} \Rightarrow a=1$. Thus, $\displaystyle 1_G\phi = 1_H$ for $\displaystyle \phi: G\rightarrow H$ a group homomorphism.
The same holds for linear maps, etc.
Just go through the proof I gave for linear maps and replace $\displaystyle +$ with $\displaystyle *$ and $\displaystyle T$ with $\displaystyle \phi$.
So it starts,
$\displaystyle \phi$ is injective
$\displaystyle \Leftrightarrow (g\phi = h\phi \Rightarrow g=h)$
$\displaystyle \Leftrightarrow ((g\phi)(h\phi)^{-1} = 1 \Rightarrow g=h)$
$\displaystyle \vdots$