Suppose that the lemma holds and that we are considering an instance of the replacement axiom, so we have a set $X$ and for every $x\in X$ there is a unique $y$ such that $\varphi(x,y,z)$, for some paramter $z$. Fix any set $w$ not in $X$, and let $R$ be the class relation such that $R(w,x)$ for each $x\in X$, and such that $R(x,y)$ whenever $\varphi(x,y,z)$. That is, $w$ points at all the members of $X$, and each member $x$ of $X$ points at it corresponding $y$. This relation is set-like, because $w$ is related to the members of $X$, which is a set, and each $x\in X$ is $R$-related only to the $y$ such that $\varphi(x,y,z)$, and this singleton is a set. But the transitive closure of $R$ will relate $w$ to all the $y$'s that arise from any $x\in X$. If the transitive closure is set-like, then the set $\{y\mid \exists x\in X\, \varphi(x,y,z)\}$ will be a set, thereby verifying this instance of the replacement axiom.

Suppose that the lemma holds and that we are considering an instance of the replacement axiom, so we have a set $X$ and for some parameter $z$ and for every $x\in X$ there is a unique $y$ such that $\varphi(x,y,z)$. Fix any set $w$ not in $X$, and let $R$ be the class relation such that $R(x,w)$ for each $x\in X$, and such that $R(y,x)$ whenever $\varphi(x,y,z)$. That is, the children of $w$ are exactly the members of $X$, and the child of any $x\in X$ is precisely the corresponding $y$. Thus, the relation $R$ is set-like, since $X$ is a set, and $\{y\}$ is a set for each $y$ that arises. But the transitive closure of $R$ will relate all the $y$'s that arise from any $x\in X$ to $w$. And so if the transitive closure of $R$ is set-like, then the set $\{y\mid \exists x\in X\, \varphi(x,y,z)\}$ will be a set, thereby verifying this instance of the replacement axiom.

Suppose that the lemma holds, that we are considering an instance of the replacement axiom, so we have a set $X$ and for every $x\in X$ there is a unique $y$ such that $\varphi(x,y,z)$, for some paramter $z$. Fix any set $w$ not in $X$, and let $R$ be the class relation such that $R(w,x)$ for each $x\in X$, and such that $R(x,y)$ whenever $\varphi(x,y,z)$. That is, $w$ points at all the members of $X$, and each member $x$ of $X$ points at it corresponding $y$. This relation is set-like, because $w$ is related to the members of $X$, which is a set, and each $x\in X$ is $R$-related only to the $y$ such that $\varphi(x,y,z)$, and this singleton is a set. But the transitive closure of $R$ will relate $w$ to all the $y$'s that arise from any $x\in X$. If the transitive closure is set-like, then the set $\{y\mid \exists x\in X\, \varphi(x,y,z)\}$ will be a set, thereby verifying this instance of the replacement axiom.

Suppose that the lemma holds and that we are considering an instance of the replacement axiom, so we have a set $X$ and for every $x\in X$ there is a unique $y$ such that $\varphi(x,y,z)$, for some paramter $z$. Fix any set $w$ not in $X$, and let $R$ be the class relation such that $R(w,x)$ for each $x\in X$, and such that $R(x,y)$ whenever $\varphi(x,y,z)$. That is, $w$ points at all the members of $X$, and each member $x$ of $X$ points at it corresponding $y$. This relation is set-like, because $w$ is related to the members of $X$, which is a set, and each $x\in X$ is $R$-related only to the $y$ such that $\varphi(x,y,z)$, and this singleton is a set. But the transitive closure of $R$ will relate $w$ to all the $y$'s that arise from any $x\in X$. If the transitive closure is set-like, then the set $\{y\mid \exists x\in X\, \varphi(x,y,z)\}$ will be a set, thereby verifying this instance of the replacement axiom.