Answer gap-filling-in topology, describing the kernel from the Seifert–van Kampen theorem The question is: 
Let $X=S^1\times I$ and let $A=S^1\times[0,3/4)$ and $B=S^1\times(1/4,1]$
So that $\{A,B\}$ is an open cover.
I have been tasked with using the the Seifert-van Kampen theorem to describe the resulting subgroup N (which refers to the kernel of $j:\pi_1(A)*\pi_1(B)\rightarrow\pi_1(X)$)
By describe it means "given an element $f\in\pi_1(A)*\pi_1(B)$ decide if it is in the kernel"

My workings:
Well $A\cap B$ is path connected (as are A and B, and X) so there's something. Truth be told I'm not sure what applying the theorem actually means.
BUT the kernel of $j$ is generated by words of the form $(i_1(g)^{-1},i_2(g))$, where $i_1:\pi_1(A\cap B)\rightarrow\pi_1(A)$ and $i_2$ is the same but goes to $\pi_1(B)$ and $g\in\pi_1(A\cap B)$ of course.
These words are members of $\pi_(A)*\pi_1(B)$
This is where I am stuck, I've tried to go "given an f, show it is in the kernel" or "given an element of the kernel, describe it" 
Thinking geometrically consider the equiv. class of a loop in $A\cap B$ if this is the identity loop (you know the one that retracts) then this will obviously be the identity in X, but that isn't ground-breaking information.
thinking with cosets doesn't really help either because I haven't explicitly found the kernel. 
I've not found an example of this and my textbook (Topology second edition, Munkres, see page 431) doesn't either. I can see (again geometrically) that the kernel is not just the identity equiv. classes (consider a clockwise loop on the upper half and an anticlockwise on the bottom, these are also "identity" loops.
 A: OK. You know that $\pi_1(X) = \mathbb Z = \pi_1(A) = \pi_1(B)$. In each case, a generator is the homotopy class of a circle of the form $\gamma(t) = (2\pi t, 1/2)$. Let's call that element $q$. Now in the free product of $\pi_1(A) * \pi_1(B)$, you have all things of the form
$$
a_1 b_1 a_2 b_2 \ldots a_k b_k
$$
where the names are meant to be suggestive, and only $a_1$ and/or $b_k$ are allowed to be the identity.  But in your case, each of these elements can be written $q^s$ for some integer $s$, whether on the "$A$" or the "$B$" side. 
Short answer: the kernel of the map consists of all things of the form
$$
q^{s_1} r^{t_1} \ldots q^{s_k} r^{t_k}
$$
(where $q$ and $r$ both denote the homotopgy class of the loop $\gamma$, but once considered as a generator of $\pi_1(A)$ and once as a generator of $\pi_1(B)$) in which the sum $s_1 + t_1 + \ldots + s_k + t_k$ is zero. 
Details: Let's let $q$ denote $[\gamma] \in \pi(X)$ (I'm going to write $\pi$ instead of $\pi_1$ from now on). Let's let $a$ denote $[\gamma]$ in $\pi(A)$, and similarly let $b$ denote it in $\pi(B)$. And let's let $t$ denote it as an element of $\pi(Y)$, where $Y = A \cap B$. 
The map $i_1$ is then 
$$
i_1 : \pi(Y) \to \pi(X) : t^k \mapsto a^k
$$
and the map $i_2$ is
$$
i_2: \pi(Y) \to \pi(X) : t^k \mapsto b^k
$$
Let's look at elements of the form $(i_1(u)^{-1}, i_2(u))$. Letting $u$ be $t^k$ for some integer $k$, these are elements like
$$
w_k = a^{-k} b^k
$$
Notice that the exponent sum in this element is zero. And so the exponent sum in any product or power of these elements will be zero as well. In particular, the inverse of $w_k$ is 
$$
v_k = b^{-k} a^k,
$$
which will be convenient to have around. 
Now I have to show that if I have a sequence of nonzero powers of alternating $a$s and $b$s, it can be written as a product of $w_k$s, or, simpler, $w_k$s and $v_k$s, for then I'll have shown that every element of the kernel is a word in $a$ and $b$ with exponent sum 0, as I claimed. 
I'm going to do this last part by example, because I'm lazy. So let's look at
$$
a^3 b^{-1} a^2 b^{-3}a^{-1}
$$
I'm going to just follow my nose here and rewrite that sequence:
\begin{align}
a^3 b^{-1} a^2 b^{-3}a^{-1} 
&=a^3 (b^{-1}) a^2 b^{-3}a^{-1}\\
&=a^3 [b^{-3} b^{2}] (a^2) b^{-3}a^{-1}\\
&=a^3 b^{-3} b^{2} [a^{-2} a^4] (b^{-3})a^{-1}\\
&=a^3 b^{-3} b^{2} a^{-2} a^4 [b^{-4} b^1] a^{-1}\\
&=(a^3 b^{-3})( b^{2} a^{-2})( a^4 b^{-4})( b^1 a^{-1})\\
&=w_{-3}     v_{-2}       w_4        v_{-1}.
\end{align}
In this rewriting, each parenthesized term in one row is replaced by the bracketed term in the next, except int he last row, where each parenthesized pair is replaced with a $v_k$ or $w_k$. 
Clearly an approach like this works in general: if you have $a^i$, you put $b^i b^{-i}$ next to it, regroup, and proceed. If that's not clear, I suppose I can write out the details, but I'm guessing that it is. 
