I want to show that $\pi_1(X,x_0)$ is a group.
I am told that $e(t) := x_0$ is the identity element.
Now, I am struggling to show that it is an identity element, and also that the inverse of an element gives $e$.
I feel like the obvious choice in defining a homotopy between $f\cdot e$ and $f$ (for some path $f : [0,1] \mapsto X$) would be,
$F(s,t) := \begin{cases} f(\frac{2}{1+s}t) , \space 0\leq t \leq \frac{1+s}{2}\\ x_0 ,\space \frac{1+s}{2} \leq t \leq 1\\ \end{cases} $
And likewise the obvious choice for defining a homotopy between $e$ and $f\cdot f^{-1}$ would be,
$G(s,t) := \begin{cases} f(2ts) , \space 0\leq t \leq \frac{1}{2}\\ g((2-2t)s) ,\space \frac{1}{2} \leq t \leq 1\\ \end{cases} $.
But I can't prove that $F$ and $G$ are continuous. So firstly am I on the right line? I.e are these the right maps to be looking at. Secondly: If so, why is it that they are continuous?
I hope you can shed some light! Thanks!
Edit :
If I can prove the following then I would be done.
I want to show that any function $H : X\times Y \mapsto Z$ is continuous if $H_x(y) := H(x,y)$ is continuous for each $x \in X$ and $H_y(x) := H(x,y)$ is continuous for each $y \in Y$. But I can't prove this either. Neither do I even know whether it is true!
Let $$A=\left\{(s,t)\in [0,1]^2\mid s\in[0,1], 0 \leq t\leq \frac{1+s}{2} \right\}$$ and let $$B=\left\{(s,t)\in [0,1]^2\mid s\in[0,1], \frac{1+s}{2} \leq t\leq 1 \right\}.$$
Note that both are closed subsets of the unit square with non-trivial intersection (given by a diagonal interval from $(0,\frac{1}{2})$ to $(1,1)$), and union the entire unit square.
$F|_A$ (the restriction of $F$ to $A$) is continuous as $f$ is continuous. $F|_B$ is also continuous because it is a constant function. We also have that $F|_A$ agrees with $F|_B$ on the intersection of $A$ and $B$ and so we may use the gluing lemma to define the unique continuous map which restricts to $F|_A$ on $A$ and $F|_B$ on $B$. This map is by definition $F$, and so $F$ is continuous.