# Pullbacks and Inclusions of Smooth function algebras of manifolds.

Let $M$ and $N$ be two smooth finite dimensional manifolds and $C^\infty(M)$ as well as $C^\infty(N)$ their smooth function algebras.

Is the following true:

Let $\imath: M \to N$ be an embedding. Then the algebra morphism $\imath^* : C^\infty (N) \to C^\infty (M)$ defined by $\imath^*(f)(m) = f(\imath(m))$ for all $m \in M$ and $f \in C^\infty(N)$ is surjective.

Let $\pi: N \to M$ be a surjective submersion. Then the algebra morphism $\pi^* : C^\infty (M) \to C^\infty (N)$ defined by $\pi^*(f)(n) = f(\pi(n))$ for all $n \in N$ and $f \in C^\infty(M)$ is injective.

• Mark, while this is a good question, it is not really research level. The second question is essentially trivial, while the first question is asking if all smooth functions on an embedded submanifold can be extended to the ambient manifold. I suggest looking at a differential geometry book such as Lee's Introduction to Smooth Manifolds - if you can't find what you're looking for there, try math.stackexchange.com I am voting to close. – MTS Apr 26 '12 at 21:15
• One can show that any $\mathbb{R}$-algebra homomorphism $C^\infty(M)\to\mathbb{R}$ has the form $f\mapsto f(x)$ for a unique point $x\in M$. Using this one can deduce that the functor $C^\infty(-)$ gives a full and faithful contravariant embedding of smooth manifolds in $\mathbb{R}$-algebras. This does not answer the question, but it is illuminating background. – Neil Strickland Apr 27 '12 at 9:21

For your first question about embeddings, yes the map is surjective. Smooth functions on $M$ can be extended to be smooth functions on $N$. One standard proof is via the smooth Urysohn Lemma, found in most manifold theory textbooks.
• For the surjectivity question and important condition is that the image of $M$ be closed in $N$. Otherwise, if $M$ has an end, smooth functions on $M$ could have wild behavior as they approach that end and not correspond to the restriction of any smooth function from the ambient manifold $N$. Think $(0,1)$ in $\mathbb{R}$. – Igor Khavkine Apr 27 '12 at 6:58
• I'm a little confused because what Igor said seems to contradict what Ryan said since I do not assume that $i(M)$ is closed in $N$. – Mark.Neuhaus May 2 '12 at 0:38