Let $M, M_1, M_2$ be manifolds. Suppose that we have smooth maps $W_1: M_1 \rightarrow M$ and $W_2 : M_2 \rightarrow M$, and that $W_1$ is a surjective submersion.

Then in particular the pullback manifold: $$M_1 \times_{M} M_2 = \{(m_1, m_2) \in M_1 \times M_2 | W_1(m_1) = W_2(m_2)\}$$ exists. I am trying to show that the projection $P: M_1 \times_{M} M_2 \rightarrow M_2$ is a surjective submersion.

Checking that $P$ is surjective is easy enough - since $W_1$ is surjective, given $m_2 \in M_2$ there exists some $m_1 \in M_1$ such that $(m_1, m_2) \in M_1 \times_{M} M_2$, but I am having trouble showing that $P$ is a submersion.

I have tried to show that given $X \in T_{m_2}(M_2)$ there is some $\bar{X} \in T_{(m_1, m_2)}(M_1 \times_M M_2)$ such that the map induced by $P$ sends $\bar{X}$ to $X$, but this has not worked so far (I have tried defining $\bar{X}$ so that $\bar{X}(f) = X(\bar{f})$ where $\bar{f}(m_2) = f(W_1^{-1}(W_2(m_2)), m_2)$, but I don't think this is a tangent vector).

Any guidance with this problem would be appreciated.


Hint: First show that $T_{(m_1,m_2)}(M_1\times_MM_2)=\{(X_1,X_2)\in T_{m_1}M_1\times T_{m_2}M_2:T_{m_1}W_1\cdot X_1=T_{m_2}W_2\cdot X_2\}$. Now take $m_2\in M_2$, a tangent vector $X_2\in T_{m_2}M_2$, and a point $(m_1,m_2)$ in the fibered product over $m_2$. Then by construction $W_1(m_1)=W_2(m_2)$ and since $W_1$ is a surjective submersion, there is a tangent vector $X_1\in T_{m_1}M_1$ such that $T_{m_1}W_1\cdot X_1=T_{m_2}W_2\cdot X_2$. But this means that $(X_1,X_2)\in T_{(m_1,m_2)}(M_1\times_MM_2)$ and of course this is mapped to $X_2$ under the derivative of the natural projection $M_1\times_MM_2\to M_2$.

  • $\begingroup$ Thanks for the response, but I don't see why $W_2$ is a submersion. $\endgroup$ – Uzai May 20 '18 at 20:04
  • $\begingroup$ Sorry I got confused by the syymetric notation you have used and did not read the assumptions carefully enough. I have edited the answer accordingly. $\endgroup$ – Andreas Cap May 21 '18 at 9:12

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