Question: So, I have recently learnt about the [Schroder-Bernstein](https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem) theorem, and was wondering if the following predicate is true:

Given $G = (V, E), H = (W, F)$ two graphs such that $\exists f : G \to H, g : H \to G$ injective graph homomorphisms, does this imply that $G \simeq H$?.

Schroder-Bernstein tells us that $V \simeq W$, however, the additional structure given by being a graph homomorphism is not necessarily preserved.

Answer: 
If we take an infinite clique and the disjoint Union of two infinite cliques, we have injective graph homomorphisms both ways but these graphs are not isomorphic.
Here is an example of two **nonisomorphic graphs** $G=(V,E)$ and $H=(W,F)$ such that  

(1) there are **bijective homomorphisms** $f:G\to H$ and $g:H\to G,$ and  

(2) each of the graphs $G$ and $H$ is isomorphic to an **induced subgraph** of the other.

Define:
$$V=W=\mathbb Z=\{\dots,-3,-2,-1,\ 0,\ 1,\ 2,\ 3,\dots\}$$
$$E=\{\{x,x+1\}:x\ge1\}\cup\{\{x,x+2\}:x\ge1\}$$
$$F=\{\{x,x+1\}:x\ge0\}\cup\{\{x,x+2\}:x\ge1\}$$

The graph $G=(V,E)$ is not isomorphic to the graph $H=(W,F)$ because $0$ is a vertex of degree $1$ in $H$, while all vertices of $G$ have degree $0,2,3,$ or $4$.

We define bijective homomorphisms $f:G\to H$ and $g:H\to G$ by setting $f(x)=x$ and $g(x)=x+1.$

Also, each of the graphs $G$ and $H$ is isomorphic to an induced subgraph of the other; namely, $G\cong H-0$ and $H\cong G-2.$