Image of Preimage under Mapping

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Theorem

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


Then:

$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B = B \cap f \sqbrk S$


Corollary

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


Proof

As $f$ is a mapping it follows by definition that $f$ is also a relation

So we apply Image of Preimage under Relation is Subset directly:

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

But from Image of Subset under Relation is Subset of Image: Corollary 3 we also have:

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

But from Preimage of Mapping equals Domain $f$ we have that $f^{-1} \sqbrk T = S$ and so:

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

Applying Image of Subset under Relation is Subset of Image: Corollary 2 we have:

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

From Intersection is Largest Subset it follows that:

$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B \cap f \sqbrk S$


Now suppose $y \in B \cap f \sqbrk S$.

Then:

$f^{-1} \sqbrk y \subseteq f^{-1} \sqbrk B$ and $f^{-1} \sqbrk y \subseteq f^{-1} \sqbrk {f \sqbrk S}$

and in particular:

$f^{-1} \sqbrk y \subseteq f^{-1} \sqbrk B$

Applying Image of Subset under Relation is Subset of Image: Corollary 2 again, we have:

$f \sqbrk {f^{-1} \sqbrk y} \subseteq f \sqbrk {f^{-1} \sqbrk B}$

But as $f$ is many-to-one, we have that:

$f \sqbrk {f^{-1} \sqbrk y} = y$

and so:

$y \in f \sqbrk {f^{-1} \sqbrk B} = \paren {f \circ f^{-1} } \sqbrk B$

So we have that:

$B \cap f \sqbrk S \subseteq \paren {f \circ f^{-1} } \sqbrk B$

Hence the result.

$\blacksquare$


Sources