Am I right to believe that a function is injective, if some elements of the first set are mapped to some elements of the second set?
It is also possible to 4 elements of the first set, are mapped to the same element of a second set?
Is this correct?
A simple answer is much appreciate, already confused enough :)
THANK YOU
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$\begingroup$$f\colon X\to Y$ is injective if and only if:
- $x\neq y\Rightarrow f(x)\neq f(y)$, or
- If $f(x)=f(y)\Rightarrow x=y$
Intuition says that you can have a replica of $X$ in $Y$, it means for all $x\in X$ there are a $y\in Y$ which $f(x)=y$ and there are not other $x'$ with the same statement, $\therefore$ Y contains a copy of the set X
Note
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- It works for $X\to Y$, if we want something similar from $Y\to X$ it is called surjective.
- Is possible that for some $y\in Y, \not{\exists}x\in X$ such $f(x)=y$
An injective function (a.k.a one-to-one function) is a function for which every element of the range of the function corresponds to exactly one element of the domain.
What this means is that it never maps distinct elements of its domain to the same element of its codomain.
A function $f:X\rightarrow Y$ is not injective if two distinct elements $a,b\in X$ exist with $f(a)=f(b)$.
If that is not the case then the function is injective.
$\endgroup$ $\begingroup$Injective function is a relation satisfying $f(x) = f(y) \Rightarrow x=y$.
$x=y \Rightarrow f(x) = f(y)$ is the definition of partial function, so we get $f(x) = f(y) \Leftrightarrow x=y$ for injective partial function.
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