Pair graphs map maps vertices into vertices and edges on edges (or single vertex) Definitions :  Let $(X,X^0),(Y,Y^0)$ be a pair of graphs (here $X^0,Y^0$ denote the subsets of vertices) and $f : X \longmapsto Y$ be continuos map.
The definition of graph I'm using is the following : $X$ Hausdorff topological space  with $\lvert X^0 \rvert < \infty$ such that $X\backslash X^0 = \sqcup_i e_i$, with $e_i$ open in $X$ homeomorphic to open intervals and $\overline{e}_i-e_i = \{1,2\}$ (one can think of $(\overline{e}_i,e_i) \sim ([0,1],(0,1))$.
Purpose : Calculate $f_* : H_1(X) \longmapsto H_1(Y)$.
In order to investigate and solve the problem, the following assumptions are useful : $f(X^0) \subset Y^0$ and $\forall \hspace{0.1cm} i, \hspace{0.1cm} \overline{e}_i$ is mapped homeomorphically on $\overline{e}_j$ or collapsed on a vertex in $Y^0$.
Problem : In general a maps between graphs don't satisfy the above requirements. What it should be true is that those are satisfied if we allow homotopy maps or splitting (i.e adding vertices to $(X,X^0)$) operations. Is this fact true? If so, how can be proved ? Any help or reference would be appreciated.
 A: This is a long comment but maybe it helps you get in the right direction.
You may already know that if $X$ is a finite connected graph with say $e$ edges and if $t$ are the number of edges in a spanning tree of $X$ (that is, a maximal subtree) then there is an isomorphism between $H_1(X)$ and $\mathbb Z^g$ where $g = e-t$. In fact, if you choose such a tree $T$ in $X$, then for each edge $e$ not in $T$ there exists a unique cycle $C_e \subseteq T\cup e$ containing $e$, and the cycles $\gamma(X) = \{ C_e : e\in X\smallsetminus T\}$ give you representatives for $H_1(X)$.
Now, for each cycle $C_e\in \gamma(X)$, and each function $f:X\to Y$ for $Y$ another graph, you can consider the image $f(C_e)$. This image will pass through some of the exceptional edges $e'$ of $Y$ corresponding to a cycle $D_{e'}$, and when it does, wind around them a certain amount of times, call this coefficient $f_{e,e'}$, and note I am allowing for it to be zero or negative.
To be more precise, each cycle $C_e$ gives you a map $i_e : S^1\to X$ and hence by composition a map $f_e : S^1\to Y$. At the same time, a cycle $D_{e'}$ in $Y$ gives you a collapsing map $\pi_{e'} : Y\to S^1$ obtained by collapsing all but the open edge $e'$ to a point (in particular, its two vertices are now identified). This gives you a map $S^1\to S^1$ that must be multiplication by some integer $f_{e,e'}$ in homology. Then I claim that:

The map $f_* : H_1(X) \longrightarrow H_1(Y)$ is given by
$ f_*[C_e] = \sum_{e'} f_{e,e'}[D_{e'}]$.

Can you prove it?
