Infinite chain of homomorphisms between triangle-free graphs with bounded max average degree? We use $G\to H$ to denote "there exists a homomorphism from graph $G$ to $H$", i.e. exists a map $h:V(G)\to V(H)$ s.t. $(x,y)\in E(G)\implies (h(x),h(y))\in H$. Let mad$(G)$ denote the max average degree of $G$, i.e. $\textrm{mad}(G):=\max_{H\subseteq G\textrm{ a subgraph}}\frac{2|E(H)|}{|V(H)|}$.
Observe that the sequence of cliques $(K_i)_{i\in \mathbb{N}}$ is an infinite ascending chain of homomorphisms in that $K_i\to K_{i+1}$ but $K_{i+1}\not\to K_i$. I wonder if an infinite ascending chain still exists when we add degree and configuration constraints. In particular,
[1] Is it true that for any $\alpha \geq 2$, there still exists an infinite ascending chain of finite, connected graphs $(G_i\stackrel{\not\leftarrow}{\to}G_{i+1})_{i\in\mathbb{N}}$ satisfying
$$(1.1)\quad \forall i, \textrm{mad}(G_i)\leq \alpha$$
and
$$(1.2)\quad \textrm{each }G_i\textrm{ is triangle-free}?$$
Note that trees are all homomorphically equivalent to a single edge, so if $\alpha < 2$ (meaning $G_i$'s do not contain cycles as subgraphs) there does not exist such an infinite ascending chain.
Also note that an infinite descending chain $(G_i\stackrel{\not\to}{\leftarrow}G_{i+1})_i$ satisfying (1.1)+(1.2) exists for all $\alpha\geq 2$: for example let $G_i:=C_{2i+5}$ the odd cycles.
Any help would be greatly appreciated. Thanks!
 A: The answer is yes for any $\alpha \geq 2$ and no for $\alpha=2$.
Actually you can find any countable partial order as an induced suborder in the planar cubic triangle-free graphs ordered by homomorphism. This is a result of Jan Hubička and Jaroslav Nešetřil.
The result is proved as follows. First, they proved that any countable partial order exists as an induced suborder in the directed paths ordered by homomorphisms. Then, they used a substitution method to change the directed graphs into undirected graphs:

(taken from p.16 from this slide)
The substitution perserves the order by homomorphisms, namely, there exists a homomorphism between directed paths $G \rightarrow H$ iff there exists a homomorphism between the respective replaced graphs.
The reason that the substitution method works is because the replaced graph (a subdivided $K_4$) is "rigid", i.e. the only homomorphism to itself is the identity.
By lemma 2.3 of the paper Chromatically optimal rigid graphs, the graph below is rigid ($l$ can be arbitrary large):

So by using this graph in the substitution method, we can generate a set of graphs with $\text{mad}$ bounded by any $\alpha>2$ and contains any countable partial order as an induced order.
For $\alpha=2$, notice that all the connected graphs are either trees or graphs with a single cycle. By taking the cores, they have cores either a cycle, an edge or a single vertex. It turns out that there is no infinite ascending chain.
