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I am a little bit confused about this question. Suppose we have function $f: \mathbb R \to \mathbb R$ by $ x \to x^2$.This is homomorphism because $\mathbb R$ is abelian.

I know that if $kerf=\{0\} \iff$ $f$ is $1-1$

Now, $kerf=\{ x \in \mathbb R : f(x) = 0_{\mathbb R}=0 \} $ In this case, we have just $0$ belongs to $kerf$. Therefore, $f$ is $1-1$ but we have already known that $y=f(x) $ is not one-to-one. So, my question is where is the mistake? I cant see

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    $\begingroup$ The set $\mathbb R$ with the product is not even a group! If you consider instead $G := \mathbb R \setminus \{0\}$ (hence "$0_G$" is $1$), then $f \colon G \to G$ is a homomorphism, and $\ker f = \{x \in G : f(x)=1\} = \{\pm1\}$. $\endgroup$
    – azif00
    Commented Jan 13, 2022 at 7:37
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    $\begingroup$ "Homomorphism" can mean different things in different contexts, so you should explain which one you have in mind, and tag the question accordingly. In particular, "$\,\ker f = \{ 0 \} \iff f$ bijective " is certainly not true for arbitrary real functions, and it would be between wrong and very unusual to call the zero set of a real function $\,\ker f\,$. $\endgroup$
    – dxiv
    Commented Jan 13, 2022 at 7:40
  • $\begingroup$ Thank all of you. I got it. Btw, @dxiv can you give me an example that you mentioned $\endgroup$
    – Elise9
    Commented Jan 13, 2022 at 7:43
  • $\begingroup$ @Elise9 Would $f(x) = \sin x$ count as an example? You still haven't clarified what space/structure this is all about. $\endgroup$
    – dxiv
    Commented Jan 13, 2022 at 7:46
  • $\begingroup$ Actually, I didnt think about it. My sister studied about functions(in high school) and I said her that 'Do you know kernel of function' and I told the property that I said before. Moreover, I wanted to give an example that she understand it better but I was confused when I said example. So, I want to ask in here to learn why is this not working. Now, I got it $\endgroup$
    – Elise9
    Commented Jan 13, 2022 at 7:53

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