# Intuition behind kernel is trivial implies homomorphism is injective.

Let $G$ and $H$ be groups, and let $\pi : G \rightarrow H$ be a homomorphism. If $\pi$ is injective, then $\ker(\pi) = \{e_G\}$. If my understanding is correct, this is because identities are mapped to identities in a homomorphism, and if $\pi$ is injective, there is only one element in the preimage of $\{e_H\}$, namely $e_G$.

Does such an intuition exist as to explain why: if $\ker(\pi) = \{e_G\}$, then $\pi$ is injective? I can prove this symbolically, but I feel like I have no clue why this should be true.

• A map of sets is injective if and only if the preimage of every element in the image consists of a single element. What is special about group homomorphisms is that the preimage of every element in the image has the same size, because all of the preimages are cosets of the kernel; the existence of inverses plays a crucial role here. – Qiaochu Yuan Nov 13 '17 at 7:29
• Oh wow. I had no idea the preimages are the cosets of the kernel. So if the kernel is trivial, we know every coset is a singleton (since cosets are the same size as their subgroup), and if the preimage of every element in the image of $\pi$ is a singleton, $\pi$ is injective? – user4396386 Nov 13 '17 at 8:05
• Yep, that's it exactly. – Qiaochu Yuan Nov 13 '17 at 18:12

It should not be surprising that if a homomorphism $\phi$ is injective, then its kernel is trivial. After all, injectivity requires that the preimage of every element of the image be unique. (And the homomorphism property requires that the preimage of the identity be the identity, in particular.)

It is the converse -- that the triviality of the kernel is sufficient for injectivity -- that is less obvious. Mark Bennett has given the core idea: symbolically, if $\phi(a)$ equals $\phi(b)$, then their (group) inverses are equal too, and thus

$$\phi(ab^{-1})=\phi(a)\phi(b^{-1})=\phi(a)\phi(b)^{-1}=e$$

The first and second steps of this calculation work only because of the homomorphism property of $\phi$. Now, if the kernel is trivial, we conclude that $ab^{-1}=e$, so $a=b$, and injectivity follows.

Here is a nonstandard but alternative way to think about this result; it is good for intuition and illustrates another Big Idea. Say a function $f$ is injective at $y$, an element of the image, if the preimage $f^{-1}(y)$ has only one element. This is a "local" property; a function might be injective at $y_1$ but not at $y_2$.

For general functions between sets, injectivity at a point is not sufficient to deduce global injectivity -- that is, injectivity at every point. But if $f$ is a group homomorphism, not just a bare function between sets, then a special type of local injectivity, namely injectivity at the identity element, is sufficient to deduce global injectivity. The extra structure provided by the homomorphism property and the definition of the identity allow us to parlay a local phenomenon into a global one.

This is a taste of a general class of "local-to-global" results that show how local phenomena can be used to deduce global structure.

If two elements of $G$ map to the same thing, their ratio maps to the identity, and is therefore in the Kernel. This ratio is only $1$ (the identity) if they are equal.

• Can you explain some more what is meant with "ratio" in this context? – Max Oct 24 at 13:25
• @Max $ab^{-1}$ would be one way of writing it - I was trying to write an intuitive formulation rather than a technical one. – Mark Bennet Oct 24 at 13:27