Show two path homotopy classes are equal? I am trying to show that $[q \circ \lambda]= [q \circ \mu]$ where $\lambda$ and $\mu$ are the paths in $I × I$ defined by
\begin{equation}
    \lambda(t)=
    \begin{cases}
      (2t,0), &  0 \leq t \leq 1/2 \\
      (1,2t-1), & 1/2 \leq t \leq 1
    \end{cases}
  \end{equation}
and 
\begin{equation}
    \mu(t)=
    \begin{cases}
      (0,2t), &  0 \leq t \leq 1/2 \\
      (2t-1,1), & 1/2 \leq t \leq 1
    \end{cases}
  \end{equation}
and $q : I × I → (I × I)/∼ $ is the quotient map to the Klein bottle.
I don't think it should be too difficult but am a little confused about the details.
 A: There is a concept of a "straight line homotopy" between maps to a convex space, which you can use to show that $\lambda$ and $\mu$ are homotopic relative endpoints as maps to $I\times I$. From this it follows that $q\circ \lambda \sim q\circ\mu$.
Say the space $Y \subset \mathbb{R}^n$ is convex if for every two points $y_0, y_1\in Y$ the straight line between them is entirely contained in $Y$. More explicitly, for all $s\in I$ the convex combination $(1-s)y_0 + sy_1$ remains in $Y$. (You should convince yourself that the straight line between $y_0$ and $y_1$ really is parametrized this way).
Now, given two continuous functions $f, g\colon X \to Y$ where $Y\subset \mathbb{R}^n$ is convex, we can define the straight line homotopy
$$ H(x, s) = (1-s)f(x) + sg(x) $$
which is continuous because it is a linear combination of continuous functions, and its image is contained in $Y$ by convexity. Moreover, this homotopy is constant on every $x$ such that $f(x) = g(x)$. In particular, if $f$ and $g$ are paths with the same start- and end-points then $H$ is a homotopy fixing end-points.
Caution: There are many cases where the straight line homotopy is not the correct thing to consider. In particular if the space $Y$ you're mapping into isn't convex then the homotopy won't necessarily stay in the space, and if $Y$ isn't a subspace of a vector space then the convex combination doesn't even make sense. It really depends on the situation and what you're trying to prove.
