# Kernel of Function

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

• 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 $\ker f = \{x \in G : f(x)=1\} = \{\pm1\}$. Commented Jan 13, 2022 at 7:37
• "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 $\,\ker f\,$.
– dxiv
Commented Jan 13, 2022 at 7:40
• Thank all of you. I got it. Btw, @dxiv can you give me an example that you mentioned Commented Jan 13, 2022 at 7:43
• @Elise9 Would $f(x) = \sin x$ count as an example? You still haven't clarified what space/structure this is all about.
– dxiv
Commented Jan 13, 2022 at 7:46
• 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 Commented Jan 13, 2022 at 7:53