Let $M$ be a (smooth) manifold and $p \in M$. We define $C^\infty_p(M)$ to be the set of equivalence classes of smooth functions $M \to \mathbb{R}$ that are identified when they agree on some neighborhood of $p$.
In Tu's book "Introduction to manifolds", p87, I read that if $U$ is an open subset of $M$ containing $p$, then $$C_p^\infty(M) = C_p^\infty(U)$$
Questions: Evidently, this is not a strict equality because on the right we have (classes of) functions $U \to \mathbb{R}$ and on the left we have functions $M \to \mathbb{R}$. Hence, I guess that they mean that there is a canonical isomorphism (of $\mathbb{R}$-algebras) $$C_p^\infty(M) \to C_p^\infty(U): f \mapsto f\vert_U$$
Clearly this is injective, because if $f$ and $g$ agree on $U$, then they are in the same class so $f= g$ in $C_p^\infty(M)$. Why is this mapping surjective? Don't we need that a smooth function $U \to \mathbb{R}$ can be smoothly extended to a smooth function $M \to \mathbb{R}$ for this?
You are right. However, germs are local, so pick a chart inside $U$ containing $p$, so that via a diffeomorphism with euclidean space you obtain a bump function $b : M \to \Bbb R$. Multiplying $f$ with $b$ may not be equal to $b$, but $f$ and $bf$ coincide near $p$ and thus they are equal as germs.