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: 
It's true for **finite** graphs. Otherwise, counterexamples abound. For example, let $G$ be a graph with $\aleph\_0$ connected components, each of which is a copy of $K\_2$; and let $H$ be a graph obtained by adding an isolated vertex to $G$. The graphs $G$ and $H$ are not isomorphic, but each is isomorphic to an induced subgraph of the other.
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 $H\cong G-2.$