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I know that for $f\colon X \to Y$, where $e_Y$ is the identity of $Y$:

$$ \ker(f) = \left\{x \in X \, \middle| \, f(x) = e_Y \right\} $$

I've learnt that kernels imply how much a homomorphism fails to be injective. But why is a function non-injective iff $|\ker(f)| > 1$? I can naively think of functions that don't work accordingly.

Most answers answer

$$|\ker(f)| > 1 \Rightarrow \text{f is non-injective}$$ which seems trivial to me, but I'm much more interested in

$$\text{f is non-injective} \Rightarrow |\ker(f)| > 1$$

i.e. why is there no such case, where only one element is mapped to identity, but the other elements have more elements mapped to them. Let's say a group homomorphism $f\colon \mathbb{Q} \to \mathbb{Z}$, both closed under addition, where $\forall x,n \in \mathbb{Z}. f(n) = f(\frac{x}{n}) = n$. It seems to me that $|\ker(f)| = 1$, however $f$ fails to be injective.

Masacroso
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hgiesel
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3 Answers3

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Assume $X,Y$ are Algebraic objects where inverting makes sense and $f$ is a homomorphism. Assume there are

$a,b \in X ; a \neq b$ with $f(a)=f(b)=c$. Then $f(ab^{-1})=f(a)f(b^{-1})=cc^{-1}=e_Y$ So $ab^{-1}$ is a non-identity element mapping to the identity. Conversely, if $|Kerf|>1$ then there are (at least) two elements mapping to $e_Y$, so the map is not injective.

gary
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If $|\mathrm{Ker}(f)|>1$, then there exists $x, y \in \mathrm{Ker}(f)$ with $x \neq y$. Then $f(x) = e_Y = f(y)$, so $f$ is not injective.

Conversely, since $f$ is a homomorphism, it preserves the algebraic structure from $X$ to $Y$, e.g., if $(X, *_X)$ and $(Y, *_Y)$ are groups, then $f(x *_X y) = f(x) *_Y f(y)$. So if $f$ is not injective, then there exists $x, y \in X$ with $x \neq y$ but $f(x) = f(y)$. Then, $e_Y = f(x) *_Y f(y)^{-1} = f(x) *_Y f(y^{-1}) = f(x *_X y^{-1}) = f(e_X)$. So $e_X, x *_X y^{-1} \in \mathrm{Ker}(f)$ and $e_X \neq x *_X y^{-1}$ by assumption.

A. S.
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I'm going to assume based on the language that you're using that you're looking at group homomorphisms.

Then if $f$ is not injective, you have $x,y\in X$ where $f(x)=f(y)$ but $x\ne y$. Then the following is true:

$$\begin{align} f(x)&=f(y)\\ \implies f(x)(f(y))^{-1}&=e_{_Y}\\ \implies f(x)f(y^{-1})&=e_{_Y}\\ \implies f(xy^{-1})&=e_{_Y} \end{align}$$

Now if $xy^{-1}=e_{_X}$, then $x=y$ (which we know not to be the case), so $xy^{-1}\in\ker(f)$ and $xy^{-1}\ne e_{_X}$, so $\lvert \ker(f)\rvert >1$.

As for your example (which actually is only defined on $\mathbb{Z}/\{0\}$), notice that you're trying to define a function of $n$, which means that $x$ must be fixed. In this case, as long as $x\ne 0$, this function will actually be injective (where it's defined) and if $x= 0$ then $\ker(f)=\mathbb{Z}/\{0\}$.

rnrstopstraffic
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