Let $G\_i$ be sequence of groups for $i\in \mathbb N$ and Let $\phi\_i$ be a monomorphism from $G\_i$ to $G\_{i+1}$.


Let $\Sigma$ be the direcet limits of $G\_i$ under the embeddings of $\phi\_i$.


Let $\varphi\_i$ be another monomorphism from $G\_i$ to $G\_{i+1}$ s.t. 


$$\phi\_i(G\_i)=\varphi\_i(G\_i)$$


i.e their image are same. (but not element wise). 


If $\Sigma '$ is the direct limits of $G\_i$ under the embeding of $\varphi\_i$ then can we say that $\Sigma\cong \Sigma '$ ?


**Note:** I had asked this question [there](https://math.stackexchange.com/questions/1168923/about-direct-limit-of-groups).



About direct limit of groups
Here is an example where they are not isomorphic (where the $G\_i$ are countable abelian groups).


Write $C\_k$ for the cyclic group of order $k$ and $C\_k^{(I)}$ the group of finitely supported functions $I\to C\_k$. 


Let $I,J$ be two disjoint infinite countable sets, and define $G=C\_2^{(I)}\oplus C\_4^{(J)}$, here $C\_2^{(I)}$ has a basis $(e\_i)\_{i\in I}$ as a free $\mathbf{Z}/2\mathbf{Z}$-module and $C\_4^{(J)}$ has a basis $(e\_j)\_{j\in J}$ and $(e\_j)\_{j\in K}$ as a free $\mathbf{Z}/4\mathbf{Z}$-module. Let $j\_0\subset J$ be some distinguished element. Let $H$ be the subgroup of $G$ (of index 2) generated by the $e\_i$ for $i\in I$, $e\_j$ for $j\in J\smallsetminus\{j\_0\}$, and $e\_{j\_0}^2$.


Let us define $G\_n=G$ (I avoid to write $G\_i$). We will construct two isomorphisms $\phi,\varphi:G\to H$, which are thus injective endomorphisms of $G$ with the same image, and define $\phi=\phi\_n$, $\varphi=\varphi\_n$ for all $n$.


Fix a bijection $c:J\to J\smallsetminus \{j\_0\}$ and an element $i\_0\in I$. Define $\phi(e\_{i\_0})=\varphi(e\_{i\_0})=e\_{j\_0}^2$, and $\phi(e\_j)=\varphi(e\_j)=e\_{c(j)}$ for all $j\in J$. 


It remains to define $\phi$ and $\varphi$ on $C\_2^{(I\smallsetminus \{i\_0\})}$, using two distinct bijections $u,v:I\smallsetminus\{i\_0\}\to I$ with suitable requirements. 


With no requirement, define $\phi(e\_i)=e\_{u(i)}$ and $\varphi(e\_i)=e\_{v(i)}$ for all $i\in I\smallsetminus\{i\_0\}$: then $\phi$ and $\varphi$ are isomorphisms $G\to H$.


For $\varphi$ we only make the requirement that $v$ has a fixed point $i\_1$ in $I\smallsetminus \{i\_0\}$, this can obviously be realized.


For $\phi$ we wish that every element of $I$ is eventually mapped to $i\_0$. For this, we write $I=\{i\_n:n\ge 0\}$ and define $u(i\_n)=i\_{n-1}$ for all $n\ge 1$. Then $u$ is indeed a bijection from $I\smallsetminus \{i\_0\}$ to $I$, and clearly any iterate of any point eventually lands in $\{i\_0\}$.


Let now $\Sigma\_\phi$ and $\Sigma\_\varphi$ be the direct limits. Then $\Sigma\_\phi$ has the property that every element of order 2 is a square, unlike $\Sigma\_\varphi$. This is because for every element of order 2 $x$ in $G$ there exists $n$ such that $\phi^n(x)$ belongs to $C\_4^{(J)}$, and hence is a square, while there exists $x$ (namely $x=e\_{i\_1}$) in $G$, of order 2, not a square, and fixed by $\varphi$, so $x$ is not a square in $\Sigma\_\varphi$.




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Edit: Here's a small variant where the direct limits are the same groups (most likely $C\_4^{(\mathbf{N})}$ and $C\_4^{(\mathbf{N})}\times C\_2^{(I)}$ for some nonempty $I$), but where the $G\_n$ are finite:


pick $G\_n=C\_2^n\times C\_4^n$, with "basis" $e\_1,\dots,e\_n,f\_1,\dots,f\_n$. Define $H\_n$ to be the subgroup of $G\_{n+1}$ generated by $e\_1,\dots,e\_{n-1},f\_{n+1}^2,f\_1,\dots,f\_n$. 


We define $\varphi\_n(e\_i)=e\_i$ for $1\le i\le n-1$, $\varphi\_n(e\_n)=f\_{n+1}^2$, $\varphi(f\_i)=f\_i$ for $1\le i\le n$.


We define $\phi\_n(e\_i)=e\_{i-1}$ for $2\le i\le n$, $\varphi\_n(e\_1)=f\_{n+1}^2$, $\varphi(f\_i)=f\_i$ for $1\le i\le n$.


Then it is not hard to check that $\Sigma\_\phi$ is isomorphic to $C\_4^{(\mathbf{N})}$ (with basis $(f\_i)\_{i\ge 1}$) and $\Sigma\_\varphi$ is isomorphic to $ C\_2^{(\mathbf{N})}\times C\_4^{(\mathbf{N})}$, with "basis" $(e\_i)\_{i\ge 1}$ and $(f\_i)\_{i\ge 1}$, where $e\_i$ is by definition the image of $e\_i$ from $G\_n$ when $n>i$.


