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Theorem 15.1 in Classical Descriptive Set Theory by Kechris states:

(i) Let $X, Y$ be Polish spaces and $f:X\rightarrow Y$ be continuous. If $A\subseteq X$ is Borel and $f|_A$ is injective, then $f(A)$ is Borel.

I assume that the last statement means that $f(A)$ is in the Borel sigma algebra on $Y$.

(ii) Now in the following corollary he states that $f$ is a Borel isomorphism of $A$ with $f(A)$.

If I now assume that $f$ is an injective function, then it is also injective on every subset of $X$. Also (ii) tells us that it is a Borel isomorphism between $X$ and $f(X)\subseteq Y$. I assume that this is meant to say that we restrict the co-domain of $f$ to its range and then this new $f$ will be bi-measurable with respect to the Borel sigma algebras on $X$ and on $f(X)$ (not on $Y$). Is that correct so far?

Now my confusion comes from the very first statement in this question (i), that for any $A\in\mathcal{B}(X)$, $f(A)\in\mathcal{B}(Y)$. It seems to me that statement (ii) should actually suggest that $f(A)\in\mathcal{B}(f(X))$?

Can maybe someone explain this to me in more detail?

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    $\begingroup$ $\mathcal B(f(X))=\mathcal B(Y)\mid f(X)=\{f(X)\cap B\mid B\in\mathcal B(Y)$ by the remark at the beginning of 11.A, so, since $f(X)\in B(Y)$, we have $\mathcal B(f(X))\subseteq\mathcal B(Y)$. $\endgroup$ Commented Nov 5 at 11:18
  • $\begingroup$ @AlessandroCodenotti thanks for your comment! So does that mean though that the statement (i) could be made more precise to say that $f(A)$ is actually in the borel sigma algebra on $f(X))$? $\endgroup$ Commented Nov 5 at 11:40
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    $\begingroup$ Yes. But in most applications you care about $f(A)$ being in $\mathcal B(Y)$ which is why the result is stated this way I'd guess $\endgroup$ Commented Nov 5 at 12:40
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    $\begingroup$ Every element $B'$ of $\mathcal B(f(X))$ is of the form $B'=f(X)\cap B$ for some $B\in \mathcal B(Y)$. If $f$ is injective then $f(X)\in\mathcal B(Y)$ and so $B'\in\mathcal B(Y)$, since $\mathcal B(Y)$ is closed under finite intersections. $\endgroup$ Commented Nov 5 at 12:56
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    $\begingroup$ No, if $B\not\in\mathcal A$ there is no reason for $\mathcal A\mid B$ to be contained in $\mathcal A$. In particular if $B$ is Borel (in $Y$) but $f(X)$ is not, then $f(X)\cap B$ will be Borel in $f(X)$ but it doesn't have to be Borel in $Y$ $\endgroup$ Commented Nov 7 at 10:12

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