Model-homogeneity strengthening: is amalgamation really needed? I'm reading some course notes on AECs (Abstract Elementary Classes) which I found on the internet, but I feel like I'm missing some obvious reason by which amalgamation is needed in an exercise.
Let $K$ be our AEC. We call $M \in K$ $\lambda$-model-homogeneous if given $N \in K_{< \lambda}$ and $M_0 \leq_K N, M$ we have some map ($K$-embedding) $N \to M$ over $M_0$.
There's a (seemingly) simple exercise asking to prove under amalgamation that we can take $N$ to have size $\leq \lambda$ instead, provided $\lambda > \operatorname{LS}{(K)}$. I just don't get why one would need amalgamation for this. AFAIK one can use "isomorphic replacements" so that model-homogeneity considers $K$-embeddings instead of simply $K$-substructures in its statement, and then we can prove it with the obvious construction of compatible maps from a chain (which gives $N$ and contains $M_0$) to $M$ using Low. Skolem, Tarski-Vaught (the chain axiom) and the original statement of model-homogeneity to construct these compatible embeddings.
Am I missing or implicitly using amalgamation somewhere? Initially I thought one needed amalgamation to pass from the $K$-embeddings to the inclusion case, but this seems something superfluous now.

EDIT:
Here's what I mean by "isomorphic replacements" in a little more detail: given some $f: M \to N$ we let $M^\ast$ be $N$ except we replace $f(M)$ with $M$, so that $f$ factors through an inclusion of $M$ in $M^\ast$, and we define an isomorphism $M^\ast \cong N$ which's $f$ on $M$ and the identity of $N$ elsewhere. By closedness by isomorphisms and iso-stability of $\leq_K$, we have that $M^\ast \in K$ and $M \leq_K M^\ast$.
Now, consider the following diagram using these isomorphic replacements, which represents the argument I had in mind and, as far as I know, doesn't rely on amalgamation:

I'd like to know what is wrong with that argument (or if I'm using AP implicitly somewhere).
 A: You have to somehow deal with the fact that model homogeneity is phrased in terms of substructures, rather than embeddings, but your idea works to do this without the amalgamation property. Let's make a definition to make this precise.
Definition. Let $K$ be an AEC. Call $M \in K$ $\lambda$-model-embedding-homogeneous if given $N \in K_{< \lambda}$ and a span of $K$-embeddings $N \xleftarrow{f} M_0 \xrightarrow{g} M$, we have some $K$-embedding $h: N \to M$ such that $hf = g$.
Clearly, $\lambda$-model-embedding-homogeneity implies $\lambda$-model-homogeneity, as the latter is just the special case where $f$ and $g$ are inclusions. The converse is also true, and can be proved using your ideas, with some tweaking.
Lemma. Let $f: A \to B$ be a $K$-embedding. Then there is $B_f \in K$ such that $A \leq_K B_f$ and an isomorphism $\theta_f: B_f \to B$, such that $f$ factors as $A \leq_K B_f \xrightarrow{\theta_f} B$.
This is essentially your idea where we 'replace' $f(A)$ by $A$ in $B$. However, that does not quite work, because the underlying sets may give trouble. For example, it could be that $A$ and $B$ have the same underlying set.
Proof. We let $B_f$ have underlying set $A$ together with a set $X$ of new elements that has the same cardinality as $B - f(A)$. Now we make $B_f$ into a structure in the obvious way, which immediately also yields the required isomorphism $\theta_f$.
Something that is not immediately obvious from the definition, but can be proved using the above lemma is the following.
Lemma. If $M' \cong M$ and $M$ is $\lambda$-model-homogeneous then so is $M'$.
Proof. Let $M_0 \leq_K M', N$ with $N \in K_{< \lambda}$. Composing the inclusion with the isomorphism $M' \cong M$ gives a $K$-embedding $f: M_0 \to M$. Now set $M_1 = f(M_0)$, then $M_1 \in K$ as $M_1 \cong M_0$. Write $g: M_1 \to N$ for the composition $M_1 \cong M_0 \leq_K N$. This allows us to build $N_g$ with $M_1 \leq_K N_g$. Now apply $\lambda$-model-homogeneity to $M_1 \leq N_g, M$ to find $h: N_g \to M$ over $M_1$. Then $N \xrightarrow{\theta_g^{-1}} N_g \xrightarrow{h} M \cong M'$ is the required $K$-embedding over $M_0$.
Proposition. A structure $M$ is $\lambda$-model-homogeneous iff it is $\lambda$-embedding-model-homogeneous.
Proof. The right to left direction is trivial, we prove the converse. Let $f,g$ be $K$-embeddings as in the definition. Consider $N_f$ and $M_g$, so that $M_0 \leq_K N_f, M_g$. As $M \cong M_g$ we have by the previous lemma that $M_g$ is $\lambda$-model-homogeneous. We thus find $h: N_f \to M_g$ over $M_0$. The composition $h': N \xrightarrow{\theta_f^{-1}} N_f \xrightarrow{h} M_g \xrightarrow{\theta_g} M$ is now the required $K$-embedding such that $h'f = g$.
After that you are right that from $\lambda$-model-embedding-homogeneity we can prove, without further use of the amalgamation property and using the exact strategy that you describe, that $N$ can be taken to have size $\leq \lambda$.

A little note about the above terminology: "$\lambda$-model-embedding-homogeneous" is not an established term, I made this up on the spot. In category-theoretic contexts (so when we view $K$ as a category) we would just call this $\lambda$-saturated, see for example Definition 2 in "Accessible categories, saturation and categoricity" by J. Rosický in The Journal of Symbolic Logic 62(3):891-901, September 1997.
Another little note that is not so mathematical in nature but more so my personal opinion: I do think that the above is a good reason as to why "$\lambda$-model-embedding-homogeneous" is a better definition, because the restriction to actual inclusions only seems to introduce annoying subtleties like the above. However, I would be very happy to hear it if someone has a counterpoint to this!
