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- #1

- Apr 14, 2013

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let $A\neq \emptyset\neq B$ be sets, $C\subseteq A$, $D\subseteq B$ subsets and $f:A\rightarrow B$ a map.

I want to show that the set $\{f^{-1}(\{x\})\mid x\in \text{im} f\}$ is a partition of $A$.

To show that the set $\{f^{-1}(\{x\})\mid x\in \text{im} f\}$ is a partition of $A$, we have to show that the union of $\{f^{-1}(\{x\})\mid x\in \text{im} f\}$ is equal to $A$, the sets $\{f^{-1}(\{x\})\mid x\in \text{im} f\}$ are disjoint and the empty set is not an element of $\{f^{-1}(\{x\})\mid x\in \text{im} f\}$, right?

I have done the following:

Since $f$ is a function, this means that two values cannot gave the same pre-image. So it follows that the sets $\{f^{-1}(\{x\})\mid x\in \text{im} f\}$ are disjoint.

Is this correct?

The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined. This means it is the set of all values $a\in A$ such that there is a $y$ with $f(a)=y$. That means that $f^{-1}(\{y\})$ contains $a$.

Does this mean that the union of all $\{f^{-1}(\{x\})\mid x\in \text{im} f\}$ is equal to $A$ ?