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Let $(\Omega, \mathcal{A})$ be a measurable space and $X:(\Omega, \mathcal{A})\rightarrow (\mathcal{X}, \mathcal{F})$ a measurable function and $f:(\mathcal{X}, \mathcal{F})\rightarrow (\mathcal{Z}, \mathcal{G})$ another measurable function.

I have stumbled across the following statement in a textbook $\sigma(f(X))\subseteq \sigma(X)$ and $\sigma(f(X))=\sigma(X)$ if and only if $f$ is one-to-one.

Now, my problem is that I dont understand if the author uses one-to-one in meaning an injective or a bijective mapping. Usually one-to-one means injection but here I have the feeling that we require $f$ to be a bijection. Can someone help?

EDIT: To add to my confusion:

$\sigma(T(X)) = \sigma(X) \iff T$ is bijective here it is being said that the statement is true if $f$ is bijective

When do we have $\sigma(X)= \sigma (f(X))$? here it is being said that the statement is true even if $f$ is only injective

But another uses suggests that the statement actually fails even when $f$ is bijective.

So what is correct now?

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  • $\begingroup$ The equality can fail even if $f$ is bijective. $\endgroup$ Commented Oct 24 at 11:16
  • $\begingroup$ See: math.stackexchange.com/q/3781898/1104384 $\endgroup$ Commented Oct 24 at 11:17
  • $\begingroup$ @KaviRamaMurthy when would this statement then be correct? Doesnt the link by Bruno B suggest that the statement is true if $f$ is bijective? $\endgroup$ Commented Oct 24 at 11:23
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    $\begingroup$ Only for standard Borel spaces. You are having general measurable spaces, so even bijection is not enough, $\endgroup$ Commented Oct 24 at 11:28
  • $\begingroup$ @KaviRamaMurthy thanks a lot! So which of the spaces would need to be standard borel and would in that case then injection or bijection be enough? $\endgroup$ Commented Oct 24 at 11:30

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