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  "body": "\n## Proof\n\n**Proof:** We give the proof assuming that *F* has no generator which is mapped to the identity of $G_1\\ast G_2$, for if there are such generators, they may be added to any of $F_1$ or $F_2$.\n\nThe following general results are used in the proof.\n\n\\1. There is a one or two dimensional CW complex, *Z* with fundamental group *F*. By Van Kampen theorem, the wedge of *n* circles is one such space.\n\n\\2. There exists a two complex $X=X_1\\cup X_2$ where $\\{p\\}=X_1\\cap X_2$ is a point on a one cell of *X* such that *X\u003csub\u003e1\u003c/sub\u003e* and *X\u003csub\u003e2\u003c/sub\u003e* are two complexes with fundamental groups *G\u003csub\u003e1\u003c/sub\u003e* and *G\u003csub\u003e2\u003c/sub\u003e* respectively. Note that by the Van Kampen theorem, this implies that the fundamental group of *X* is $G_1\\ast G_2$.\n\n\\3. There exists a map $f:Z\\rightarrow X$ such that the induced map $f_\\ast$ on the fundamental groups is same as $\\phi$\n\nFor the sake of convenience, let us denote $f^{-1}(X_1)=:Z_1$ and $f^{-1}(X_2)=:Z_2$. Since no generator of *F* maps to identity, the set $Z_1\\cap Z_2$ has no loops, for if it does, these will correspond to circles of *Z* which map to $p\\in X$, which in turn correspond to generators of *F* which go to the identity. So, the components of $Z_1\\cap Z_2$ are contractible. In the case where $Z_1\\cap Z_2$ has only one component, by Van Kampen's theorem, we are done, as in that case, :$F=\\Pi_1(Z_1)\\ast\\Pi_1(Z_2)$.\n\nThe general proof follows by reducing *Z* to a space homotopically equivalent to it, but with fewer components in $Z_1\\cap Z_2$, and thus by induction on the components of $Z_1\\cap Z_2$.\n\nSuch a reduction of *Z* is done by attaching discs along binding ties.\n\nWe call a map $\\gamma :[0,1]\\rightarrow Z$ a **binding tie** if it satisfies the following properties\n\n\\1. It is **monochromatic** i.e. $\\gamma([0,1])\\subseteq Z_1$ or $\\gamma([0,1])\\subseteq Z_2$\n\n\\2. It is a **tie** i.e. $\\gamma(0)$ and $\\gamma(1)$ lie in different components of $Z_1\\cap Z_2$.\n\n\\3. It is **null** i.e. $f \\circ \\gamma([0,1])$ is null homotopic in **X**.\n\nLet us assume that such a binding tie exists. Let $\\gamma$ be the binding tie.\n\nConsider the map $g:[0,1]\\rightarrow D^2$ given by $g(t)= e^{it}$. This map is a [[homeomorphism|0QT00PB5G2]] onto its image. Define the space $Z'$ as\n\n$Z'= Z \\coprod\\! D^2/\\! \\sim$ where :$x\\!\\!\\sim y \\text{ iff} \\begin{cases} x=y, \\mbox{ or }\\\\ x=\\gamma (t) \\text{ and } y= g(t) \\text{ for some } t\\in [0,1]\\mbox{ or }\\\\ x=g (t) \\text{ and } y= \\gamma (t) \\text{ for some } t\\in [0,1] \\end{cases}$\n\nNote that the space *Z'* deformation retracts to *Z* We first extend *f* to a function $''f'':Z\\coprod \\partial D^2/\\!\\sim$ as\n\n$$f''(x) = \\begin{cases}f(x),\\ x\\in Z\\\\ p \\text{ otherwise.}\\end{cases}$$\n\nSince the $f(\\gamma)$ is null homotopic, $f''$ further extends to the interior of the disc, and therefore, to $Z'$. Let $Z_i' = f'^{-1}(X_i)$ *i = 1,2*. As $\\gamma(0)$ and $\\gamma(1)$ lay in different components of $Z_1\\cap Z_2$, $Z_1' \\cap Z_2'$ has one less component than $Z_1\\cap Z_2$.\n",
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