# Thread: quick question on paths

1. ## quick question on paths

If $\displaystyle f: X \rightarrow Y$ is a map and if $\displaystyle g$ and $\displaystyle h$ are paths in $\displaystyle X$ with $\displaystyle g(1)=g(0)$, then $\displaystyle f(g \cdot h)=fg \cdot fh$.

I was told that this follows directly from the definition of path composition. This is where $\displaystyle f:X \rightarrow Y$ so the first is a path in $\displaystyle Y$ and the second is a path in $\displaystyle Y$. Is this completely justified by this, or do I need to justify this more than what is above?

2. Hi

$\displaystyle g(1)=h(0)$ I suppose.

Yes that equality is quite natural. If you want to write and see, well, by definition:
$\displaystyle g.h(t)=\begin{cases} g(2t) & 0\leq t\leq \frac{1}{2} \\ h(2t-1) & \frac{1}{2}\leq t\leq 1\\\end{cases}$
Therefore
$\displaystyle f(g.h(t))=\begin{cases} f(g(2t)) & 0\leq t\leq \frac{1}{2} \\ f(h(2t-1)) & \frac{1}{2}\leq t\leq 1\\\end{cases}$

If you consider the two paths $\displaystyle fg$ and $\displaystyle fh$ in $\displaystyle Y$, then by definition:

$\displaystyle fg.fh(t)=\begin{cases} fg(2t) & 0\leq t\leq \frac{1}{2} \\ fh(2t-1) & \frac{1}{2}\leq t\leq 1\\\end{cases}$

So the equality comes from the fact that $\displaystyle \forall t\in [0,1],\ f(g(t))=fg(t)\ \text{and}\ f(h(t))=fh(t),$ i.e. from the definition.