If $fmM:X\to Y$ is continuous then $fmM:X\to f(X)$ is continuous

Theorem 1.

Suppose $X\mathrm{,}Y$ are topological spaces and $f\mathrm{:}X\mathrm{\to }Y$ is a continuous function. Then $f\mathrm{:}X\mathrm{\to }f\mathit{}\mathrm{(}X\mathrm{)}$ is continuous when $f\mathit{}\mathrm{(}X\mathrm{)}$ is equipped with the subspace topology.

Proof.

Let us first note that using a property on this page (http://planetmath.org/InverseImage), we have

$$X=f^{-1}f(X).$$

For the proof, suppose that $A$ is open in $f(X)$, that is, $A=U\cap f(X)$ for some open set $U\subset Y$. From the properties of the inverse image, we have

$$f^{-1}(A)=f^{-1}(U)\cap f^{-1}(f(X))=f^{-1}(U)$$

so $f^{-1}(A)$ is open in $X$. ∎

Title	If $fmM:X\to Y$ is continuous then $fmM:X\to f(X)$ is continuous
Canonical name	IfFcolonXtoYIsContinuousThenFcolonXtoFXIsContinuous
Date of creation	2013-03-22 15:16:28
Last modified on	2013-03-22 15:16:28
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	6
Author	matte (1858)
Entry type	Theorem
Classification	msc 26A15
Classification	msc 54C05
Related topic	ContinuityIsPreservedWhenCodomainIsExtended