Image of Preimage under Mapping/Corollary

Corollary to Image of Preimage under Mapping

Let $f: S \to T$ be a mapping.

Then:

$B \subseteq \Img f \implies \paren {f \circ f^{-1} } \sqbrk B = B$

where:

$f \sqbrk X$ denotes the image of $X$ under $f$

$f^{-1} \sqbrk X$ denotes the preimage of $X$ under $f$

$f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$

$\Img f$ denotes the image set of $f$.

Proof

From Preimage of Subset is Subset of Preimage:

$B \subseteq \Img f \implies f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img f}$

and from Intersection with Subset is Subset:

$f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {\Img f} \implies f^{-1} \sqbrk B \cap f^{-1} \sqbrk {\Img f} = f^{-1} \sqbrk B$

Hence the result.

$\blacksquare$

Sources

• 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): $\S 2.2$: Homomorphisms: $\text{(i)}$