Why is the fundamental group of torus with finitely many points removed a bouquet of circles? I am looking for a full explanation for how to compute $\pi_1(S^1\times S^1 \setminus \{x_1, \ldots, x_n\})$ where each $x_i\in S^1\times S^1$. I know that this has been asked on SE before, but the answers are always brief sketches.
I think that the final answer comes out to the free product $\ast_{i=1}^{n+1}\mathbb Z_i$ where each $\mathbb Z_i=\mathbb Z$, but I have little idea as to how we arrive at this.
I would really appreciate it if someone could help me understand the full process of computing this fundamental group. It would also be great if the explanation could stick to relatively basic concepts (i.e., basic properties of free groups, Van-Kampen, etc.).
UPDATE (this is not a duplicate): 
I realize that my question was too vague. Here is some more direction. 
The answer should come out to a bouquet of $n+1$ circles, I believe. My intuitive ideas is as follows. The torus with one point removed deformation retracts onto $\mathbb Z \ast \mathbb Z$. Now, my justification for this is a little fuzzy, though. I can see that $S^1\times I$ deformation retracts onto $S^1$ where $I=[0,1]$, so that $\pi_1(S^1\times I)\cong \mathbb Z$, but I can't draw an explicit connection between this fact and the first deformation retraction. Any help seeing the explicit argument here is appreciated. 
Moreover, I don't see in general how removing more points continues to add to the bouquet of circles. Would we need an induction argument? If so, how would it go?
I  hope this clarifies my question.
 A: This can be done simply and with complete rigor with a quotient map construction and a deformation retraction construction. Van Kampen's Theorem is not needed, but you will need the theorem that the fundamental group of a graph is a free group. As with many quotient constructions, the idea for making the construction rigorous is to visualize what the picture should be, and then translate your visualization into mathematical formulas for quotient maps, subsets, etc. The comments of @QiaochuYuan give some hints how to do this, I will be somewhat more explicit.
By the way, this is hard to do without having good visualization skills combined with good skills at writing down the appropriate analytic formulas. Perhaps it is annoying to do this, but on the other hand it is satisfying in the end to be able to see the connection between the visualization and the analytic formulas, and to see that the end product is completely rigorous. 
I won't write down the formula for final deformation retraction, but I will write down formulas for all of the relevant points and subsets, which are very simple, and I will describe the visualization for the final deformation retraction.
Start with $I^2$ and its standard quotient map $q : I^2 \to T^2$.  
Instead of removing just the one point $\{1/2,1/2\}$ from $I^2$, remove the following $n$-points from $I^2$:
$$\widehat{x}_i=\{(2i−1)/2n,1/2\}, \quad i=1,...,n
$$
Under the quotient map, their images $x_i$ are removed from $T^2$. 
Instead of deformation retracting $I^2 - (1/2,1/2)$ onto the boundary of $I^2$, instead use a deformation retraction $\widehat D$ from $I^2 - \{\widehat x_1,...,\widehat x_n\}$ onto the set $\widehat \Gamma$ which is union of the boundary of $I^2$ and the vertical segments $\{i/n\} \times [0,1]$. 
Under the quotient map $q : I^2 \mapsto T^2$, the image of $\widehat\Gamma$ is a finite connected graph $\Gamma$ having $n$ vertices and $2n$ edges, hence $\chi(\Gamma) = -n$, hence $\pi_1\Gamma$ is free of rank $1 - \chi(\Gamma)=1+n$. 
Also, under the quotient map the deformation retraction $\widehat D : I^2 - \{\widehat x_1,...,\widehat x_n\} \to \widehat \Gamma$ descends to a deformation retraction $D : T^2 - \{x_1,...,x_n\} \to \Gamma$. Thus $T^2 - \{x_1,...,x_n\}$ is homotopy equivalent to $\Gamma$ and has fundamental group free of rank $1+n$.
The formula for the deformation retraction $\widehat D$ (and hence for $D$) can also be written down with a little work. Geometrically, working on $I^2$, in each subrectangle $[2i-2,2i] \times [0,1]$ you just radially project away from the point $(2i-1,1/2)$ onto the union of boundary segments of that subrectangle.  Writing down an actual formula to do this should be a simple mechanical exercise in analytic geometry.
