Intuition behind kernel is trivial implies homomorphism is injective.

Question

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.

Answer 1

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.

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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 local injectivity at even a single point is sufficient to deduce global injectivity. The extra structure provided by the homomorphism allows 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.

Answer 2

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.

Answer 3

Just note that the preimage set of a given image element is the coset of the kernel by any preimage element of the given image element. So, every preimage set is a singleton (namely the homomorphism is injective) if and only if the kernel is trivial.