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So my prof gave me this proof:

$f(x) = f(y) ⇐⇒ f(y − x) = 0 ⇐⇒ y − x ∈ Ker f.$

I dont see why this proof is enough, this only says $y-x \in Ker f$

asddf
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2 Answers2

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First suppose $f$ is injective.

Since $f$ is linear, $f(0) = 0$, hence $0 \in \text{ker}(f)$.

But if $x$ is any element of $\text{ker}(f)$, then \begin{align*} &x \in \text{ker}(f)&&\\[4pt] \implies\; &f(x) = 0&&\\[4pt] \implies\; &f(x) = f(0)&&\text{[since $f(0) = 0$]}\\[4pt] \implies\; &x = 0&&\text{[since $f$ is injective]}\\[4pt] \end{align*}

It follows that $\text{ker}(f) = \{0\}$.

Thus, $f$ injective implies $\text{ker}(f) = \{0\}$.

Next, suppose $\text{ker}(f) = \{0\}$. Then \begin{align*} &f(x)=f(y)&&\\[4pt] \implies\; &f(x)-f(y) = 0&&\\[4pt] \implies\; &f(x-y) = 0&&\text{[since $f$ is linear]}\\[4pt] \implies\; &x-y \in \text{ker}(f)&&\\[4pt] \implies\; &x-y = 0&&\text{[since $\text{ker}(f) = \{0\}$]}\\[4pt] \implies\; &x=y&&\\[4pt] \end{align*}

hence $f$ is injective.

Thus, $\text{ker}(f) = \{0\}$ implies $f$ is injective.

Hence, $f$ is injective $\iff \text{ker}(f) = \{0\}$, as was to be shown.

quasi
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  • Should be the accepted answer – The Coding Wombat Jan 24 '19 at 23:26
  • How does $f(x) = f(0)$ imply $x=0$? The definition of injectivity states $x=y \Rightarrow f(x) = f(y)$ which is not an equivalence. – rndm_me May 23 '20 at 13:06
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    @rndm_me: The definition of injectivity is $f(x)=f(y){,\implies,}x=y$, or equivalently, $x\ne y{,\implies,}f(x)\ne f(y)$. The implication $x=y{,\implies,}f(x)=f(y);$is true for any function $f$. – quasi May 23 '20 at 15:53
  • @quasi one thing is not clear to me in direct part of the proof: you put $0=f(0)$ and hence $f(x)=0\Rightarrow f(x)=f(0)\Rightarrow x=0$.See I need to proof that 0 is the only element of $Kerf$, initially I don't know what is $Kerf$,so one may think that $0=f(y)$ for some y in domain and not necessarily $0=f(0)$ and hence we get in general x=y,so how can you say that y=0 necessarily,I mean if i am not wrong you are using what you need to prove – Ibrahim Islam Jul 07 '20 at 12:51
  • @lbs: $f(0)=0$ is true since $f$ is linear. Hence $0\in\text{ker}(f)$. Next we show: If $x\in \text{ker}(f)$, then $x=0$. To show an if-then type statement, we get to assume the "if" part. Thus we get to assume $x\in\text{ker}(f)$, and our goal is to show $x=0$. From $x\in\text{ker}(f)$, we get $f(x)=0$, But we also know $f(0)=0$, hence $f(x)=0$ can be rewritten as $f(x)=f(0)$. Then since in the forward direction part of the proof, we know that $f$ is injective, from $f(x)=f(0)$, we get $x=0$. It follows that $0$ is the only element of $\text{ker}(f)$, hence $\text{ker}(f)={0}$. – quasi Jul 07 '20 at 14:38
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I dont see why this proof is enough, this only says $y-x \in Ker f$

It says much more, since you have the symbol $$ \iff $$ meaning that each side implies the other side. Note that $Ker f=\{0\}$ allows one to conclude.

Olivier Oloa
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