The definition of a homomorphism $f$ from $G$ to $H$, given by Pinter, says that: If $G$ and $H$ are groups, a homomorphism from $G$ to $H$ is a function $f: G \rightarrow H$ such that for any two elements $a,b \in G$, $f(ab)=f(a)f(b)$. If there exists a homomorphism from $G$ onto $H$, we say that $H$ is a homomorphic image of $G$.
So, I am wondering, is a homomorphism onto, or one-to-one? An explanation would be appreciated. It seems to me, that a homomorphism should be just one-to-one.
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$\begingroup$A homomorphism $f:G\to H$ need be neither 1-1 nor onto. It need merely satisfy the requirement that $f(ab)=f(a)f(b)$. Note that the product $ab$ is the product in the domain $G$, while the product $f(a)f(b)$ is the product in the range $H$. One always has that $f(G)\subset H$.
If it is 1-1, it is called a monomorphism.
If it is onto, it is called an epimorphism. This means $f(G)=H$.
If it is both 1-1 and onto, it is called an isomorphism.
$\endgroup$ $\begingroup$Here is an example to illustrate the falsity of your claim. One of them requires calculus to understand.
For surjectivity, consider the map $\phi: \mathbb{Z} \to \mathbb{Z}$ where $\Bbb Z$ is the usual integer group be defined by $\phi(x) = 2x.$ The $x = 3$ for instance, has no suitable preimage element.
For injectivity, consider the group $F$ where it is the additive group of smooth functions on $\Bbb R$. The map $\phi: F \to F$ is a homomorphism (by the additive laws of derivatives) defined by $\phi(f) = f'$ , but if we select $f'(x) = e^x$, then it is not injective because the family $f(x) = e^x + c$ maps to $f'(x) = e^x$.
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