Today, in my class of set theory, my professor said that replacement axiom doesn't true in the structure $(\omega, \in)$. However, I think that he was wrong and here is my proof.
First the axiom form that I use is the following: Let $\phi(x,y,A,w_1,\dots,w_n)$ be a formula in the language of set theory where $x,y,A,w_1,\dots,w_n$ are parameters. So, replacement axiom says: $$ \forall A,w_1,\dots,w_n(\forall x\in A\exists! y\phi\rightarrow\exists Y\forall x\in A\exists y\in Y\phi). $$
Now, let $\phi$ a formula as above and take $A,w_1,\dots,w_n\in \omega$. Also, suppose that $\forall x\in A\exists!\phi$. As $A$ is a natural number, $A=\{0,1,\dots,A-1\}$. Now, define $y_k\in\omega$ as the unique element such that $\phi(k,y_k,w_1,\dots,w_n)$ where $k\in A$.
If we take $Y=\max\{y_1,\dots,y_{A-1}\}+1$, then $\forall x\in A\exists y\in Y\phi$. So, the axiom holds.
Is there some mistake in my argument?
The professor is correct. Let $t=\{0, 1, 2\} (= 3 \in \omega)$ and let $\phi(x, y)\text{ be } y=x \cup \{x\}.$ Then $\{1, 2, 3\} \notin \omega$ (although $\{1, 2, 3\} \subset \omega$), so replacing each element of $3$ with its image under $\phi$ results in something that is not a member of your universe.
As @spaceisdarkgreen points out, this argument assumes the usual definition of Replacement. In Kunen's weaker definition, though, the axiom scheme merely states that there is a set containing (rather than consisting of exactly) the "replacement" elements. With this definition, Replacement holds (though Comprehension fails) in $(\omega, \in)$.
To see this, note that every member of $\omega$ is finite, so the "replacement" elements will also constitute a finite subset of $\omega$, which therefore has a largest element $N$. Then $N+1$ is a "witness" demonstrating this instance of Replacement.