Inverse Mapping/Examples/Arbitrary Finite Set with Itself
Example of Compositions of Mappings
Let $X = Y = \set {a, b}$.
Consider the mappings from $X$ to $Y$:
| \(\text {(1)}: \quad\) | \(\ds \map {f_1} a\) | \(=\) | \(\ds a\) | |||||||||||
| \(\ds \map {f_1} b\) | \(=\) | \(\ds b\) |
| \(\text {(2)}: \quad\) | \(\ds \map {f_2} a\) | \(=\) | \(\ds a\) | |||||||||||
| \(\ds \map {f_2} b\) | \(=\) | \(\ds a\) |
| \(\text {(3)}: \quad\) | \(\ds \map {f_3} a\) | \(=\) | \(\ds b\) | |||||||||||
| \(\ds \map {f_3} b\) | \(=\) | \(\ds b\) |
| \(\text {(4)}: \quad\) | \(\ds \map {f_4} a\) | \(=\) | \(\ds b\) | |||||||||||
| \(\ds \map {f_4} b\) | \(=\) | \(\ds a\) |
We have that:
- $f_1$ is the inverse mapping of itself
- $f_4$ is the inverse mapping of itself
- the inverse of neither $f_2$ nor $f_3$ are mappings.
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $2$: Maps and relations on sets: Example $2.13$