I have a question about he argumentation of this proof.
Problem: Given are two graphs $S$ and $T$. All the vertices in graph $S$ have an degree (number of neighbors) of $3$ and the vertices of $T$ have a degree of $2$ to $5$. For example $v_1 \in V(G)$ with $d(v_1)=4$ Is there an Isomorphism from $S$ to $T$?
The argumentation of the Proof is:
There is no Isomorphism from $S$ to $T$ because all vertices of $S$ have an degree of $3$ an and $d(v_1)=4$, so there can not be Isomorphism.
My question would be why? Is it because of the surjectivity ?
An isomorphism $f\colon T\to S$ maps $v_1$ to a vertex $f(v_1)$ of $S$ and its four neighbours $w_1,w_2,w_3,w_4$ to four distinct vertices $f(w_i)$ that are neighbours of $f(v_1)$ - but $f(v_1)$ has only three neighbours!
Rule of thumb: Isomorphic thingies are indistinguishable as long as we consider only general thingie-properties. Your graphs $S$ and $T$ can be distinguished by asking "Is there a vertex of degree $\ne 3$?"