How to find the kernel of the homomorphism between two fundamental groups. Let $X = T^2 - \{x_1\}$. 
I need to describe the kernel of the homomorphism 
$$\phi :\pi_1(X,x_0)  \rightarrow \pi_1(T^2,x_0).$$
I know $$\pi_1(X,x_0) \cong F(a,b)$$ and $$\pi_1(T^2,x_0) \cong \pi_1(S^1,x_0) \times \pi_1(S^1,x_0) \cong \mathbb{Z} \times \mathbb{Z} $$
Since $\pi_1(S^1,x_0)$ just describes the number of times a loop goes around in $S^1$ my first thought was to say
$$\begin{align} \phi(a^m) &\mapsto (m,0) \\  
\phi(b^n) &\mapsto (0,n)
\end{align}$$
So that $\phi(a^mb^n) = \phi(a^m)\phi(b^n) = (m,0)+(0,n)$
Then the kernel of $\phi$ are elements whose total power on $a$ sum to zero and whose total power on $b$ sum to zero. Is this correct or am I completely off?
 A: You need to be careful with your base points since $x_0$ is not in $X$. So pick a basepoint other than $x_0$ for both of them. Depending on how you proved that $T^2$ has fundamental group $F_2$ the following fact is more or less obvious: if $\alpha, \beta$ are representatives of the two generators of $\pi_1(T^2,y)$ then they are also representatives of the generators of $\pi_1(X,y)$.
From here your line of thinking applies, but be careful with how you phrase it: the total power of $a$ should be 0 and the total power of $b$ should be 0.
There is a more algebraic way to look at it:
By the observation above, the inclusion map $X \rightarrow T_2$ induces a surjective map on fundamental groups $F_2 \rightarrow \mathbb{Z}^2$.
Lets prove the following fact: If we have a surjective homomorphism $\psi: F_n \rightarrow \mathbb{Z}^n$, the kernel of $\psi $ is the commutator subgroup of $F_n$.
Basic group theory tells us that the kernel must contain the commutator. By the third isomorphism theorem (I don't think I've ever used this before), $\mathbb{Z}^n \cong F_n / ker(\psi) \cong (F_n /F_n') / (ker(\psi) /F_n')$. The rightmost group is a quotient group of $\mathbb{Z}^n$ that is isomorphic to $\mathbb{Z}^n$. Call the quotient map $\rho$ Look at the short exact sequence $0 \rightarrow ker(\rho) \rightarrow \mathbb{Z}^n \rightarrow \mathbb{Z}^n \rightarrow 0$. It splits because the image of $\rho$ is free. This and basic facts about free abelian groups imply that the kernel is trivial. Thus, $ker(\psi) /F_n'$ is trivial which means that the kernel is exactly $F_n'$.
Applying this fact above shows that the kernel has to be exactly the derived subgroup $F_2'$.
Note this group theory (and your correct deduction about the kernel) tells you something nontrivial, that the derived subgroup of $F_n$ is the same as the subgroup consisting of elements where each basis element is used a total of 0 times.
