Is this proof correct? I am particularly uncertain with the last step.
Consider $f: X \to Y, g: M \to N$. $\forall x \in X, df_x: T_x(X) \to T_y(Y)$ is injective.
Similarly, $\forall m \in M, dg_m: T_m(M) \to T_n(N)$ is injective.
Therefore, $\forall (x,m), x \in X, m \in M, d(f \times g)_{x,m} = df_x \times df_m$. Since both $df_x$ and $dg_m$ are injective, $d(f \times g)_{x,m} $ is injective.
I am very much lost here, not even sure what do I suppose to prove:
Do I need to show that $d(f \times g)_{x,m} = df_x \times df_m$? If so, how can I prove it?
Also,
Is the other thing I am asked to show is Since both $df_x$ and $dg_m$ are injective, $d(f \times g)_{x,m} $ is injective? If so, is the following attempt correct?
Given $d(f \times g)_{x,m}(\tilde{x}, \tilde{m}) = (0_y, 0_n)$, where $\tilde{x} \in T_x X, \tilde{m} \in T_x M$. Here I denote $0_y, 0_n$ as vector in the tangent space of $Y,N$ for clarity.
But $d(f \times g)_{x,m}(\tilde{x}, \tilde{m}) = df_x \times df_m(\tilde{x}, \tilde{m}) = df_x(\tilde{x}) \times df_m(\tilde{m}) = (0_y, 0_n)$. That says $df_x(\tilde{x}) = 0_y$, and $df_m(\tilde{m}) = 0_n$. By the injectivity of $df_x, df_m$ we know $\tilde{x} = 0_x, \tilde{m} = 0_m.$ Here I denote $0_x, 0_m$ as vector in the tangent space of $X,M$ for clarity.
Hence, $d(f \times g)_{x,m}(\tilde{x}, \tilde{m}) = (0_y, 0_n)$ implies $(\tilde{x}, \tilde{m}) = (0_x, 0_m)$. Therefore, $d(f \times g)_{x,m} $ is injective. So $f \times g$ is an immersion.