# Showing that the product of two liftings is a lift of the product of two paths.

algebraic-topologycovering-spacesgeneral-topology

I'm just learning about covering spaces, and there's a very simple exercise from Munkres I'm struggling with.

Let $p:E\to B$ be a covering map, and let $\alpha$ and $\beta$ be paths in $B$ such that $\alpha(1)=\beta(0)$; let $\tilde{\alpha}$ and $\tilde{\beta}$ be their liftings such that $\tilde{\alpha(1)} = \tilde{\beta(0)}$. Show that $\tilde{\alpha} \cdot \tilde{\beta}$ is a lifting of $\alpha\cdot \beta$.

We don't know that $\alpha\cdot \beta$ forms a loop, so I can't apply any of the "heavyweight" theorems that use the lifting correspondence so the proof must be quite elementary. The path lifting lemma guarantees that a lifting of $\alpha\cdot\beta$ exists and is unique, but I'm not sure how to use this to show that $p\circ (\tilde{\alpha} \cdot \tilde{\beta}) = \alpha \cdot \beta$.

I just figured it out, it follows trivially from the equation $h\circ(f\cdot g) = (h\circ f) \cdot (h \circ g)$