Subset of Domain is Subset of Preimage of Image/Equality does Not Necessarily Hold
Jump to navigation
Jump to search
Theorem
Let $f: S \to T$ be a mapping.
From Subset of Domain is Subset of Preimage of Image:
- $A \subseteq S \implies A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
where:
- $f \sqbrk A$ denotes the image of $A$ under $f$
- $f^{-1} \sqbrk A$ denotes the preimage of $A$ under $f$
- $f^{-1} \circ f$ denotes composition of $f^{-1}$ and $f$.
It is not necessarily the case that:
- $A \subseteq S \implies A = \paren {f^{-1} \circ f} \sqbrk A$
Proof
Let:
- $S = \set {0, 1}$
- $T = \set 2$
Let $f: S \to T$ be defined as:
- $\map f 0 = 2$
- $\map f 1 = 2$
Let $A \subseteq S$ be defined as:
- $A = \set 0$
Then we have:
- $f \sqbrk A = \set 2$
but:
- $f^{-1} \circ f \sqbrk A = f^{-1} \sqbrk 2 = \set {0, 1}$
That is:
- $A \subseteq \paren {f^{-1} \circ f} \sqbrk A$
but it is not the case that:
- $A \paren {f^{-1} \circ f} \sqbrk A$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.13 \ \text{(c)}$