Question on Axiom of replacement

Question

This axiom comes from chapter 3 "set theory" of Tao Analysis I

Axiom $3.6$ (Replacement). Let $A$ be a set. For any object $x \in A$ and any object $y$, suppose we have a statement $P(x, y)$ pertaining to $x$ and $y$, such that for each $x\in A$ there is at most one $y$ for which $P(x,y)$ is true. Then there exists a set $\{y: P(x, y) \text{ is true for some } x \in A\}$ such that for апy object $z$, $$ z\in \{y : P(x, y)\text{ is true for some } x \in A\} \iff P(x,z)\text{ is true for some } x \in A.$$

For me, this axiom implies a way to construct a new set $B$ from the original set $A$, and it's a mapping from set $A$ to set $B$. So it's obvious that for each $x\in A$ there is exactly one $y$ for which $P(x,y)$ is true.

My question is: Can I replace "at most one" with "exactly one" in the axiom?

Thank you guys, I think my question should be closed. Because I got this from here:The axiom of replacement basically says that if A is a set and f is an operation on elements of A, then $\{f(x) : x \in A\}$ is a set. Here the operation f may return an undefined result (because for each $x$, the statement $P(x,y)$ is true for at most one $y$ rather than exactly one $y$). So to construct the set $\{x \in A : P(x) \text{ is true}\}$, we can define $f(x)$ to be $x$ if $P(x)$ is true, and leave $f(x)$ undefined if $P(x)$ is false.

Answer 1

Yes they are equivalent. Using the axiom of separation you can construct a subset $\mathcal{A}\subseteq A$, such that for each $x\in \mathcal{A}$ there is exactly one $y$ for which $P(x,y)$ is true, applying your version of replacement on this set, we can prove Tao’s version.