Let $M$ be a $n$-dim manifold, $f:M\rightarrow \mathbb{R}^n$ smooth. If we write $f$ as $(f_1,...,f_n)$ and I know $$df_1|_p: T_pM \rightarrow T_{f_1(p)}\mathbb{R}$$ is the constant zero linear function. I would like to show that $df|_p$ is singular.
From definition, $df|_p :T_pM \rightarrow T_{f(p)}\mathbb{R}^n$, I claim that the vector $\frac{\partial}{\partial x_1}\bigg|_p \in T_pM $ will be sent to the zero vector in $T_{f(p)}\mathbb{R}^n$, which will show $df|_p$ is singular. To show the image is the zero vector, we need to show that for each $g:\mathbb{R}^n \rightarrow \mathbb{R}$ smooth, we have $$df|_p \left(\frac{\partial}{\partial x_1}\bigg|_p\right)(g) = 0.$$ From definition, this is $$\frac{\partial}{\partial x_1}\bigg|_p (g\circ f)$$ I was wondering how would I apply the chain rule here? I only know we have the chian rule for the differential map like $d(u \circ v)$.
Edit: with the answer below, I will try to write out the solution in detail.
Let $r_1:\mathbb{R}^n \rightarrow \mathbb{R}$ be the mapping that sends a vector in $\mathbb{R}^n$ to its first component. Note that $r_1$ can be seen as the first component of a coordinate map on $\mathbb{R}^n$, and therefore $dr_1|_{f(p)}$ can be identified as an element in $(T_{f(p)}\mathbb{R}^n)^*$. Look at the dual map $\delta f:(T_{f(p)}\mathbb{R}^n)^* \rightarrow (T_p M)^*$ from definition $$\delta f(dr_1)(\frac{\partial}{\partial x_i}) = r_1(df(\frac{\partial}{\partial x_i})) = 0 \quad i=1,2,...,n$$ therefore $\delta f$ is not injective, therefore $df$ is not surjective, but the domain and codomain of $df$ both have dimension $n$, therefore $df$ is singular.
Your derivative is ok, but does not show that $\partial_1$ is sent to zero, and it isn't. The image is the vector $$f_* \partial_1 = \left( 0, \partial_1 f_2, ... , \partial_1 f_n\right) \ \ \mbox{or} \ \ f_*\partial_1 = \sum_{k\geq 2} \frac{\partial f_k}{\partial x_1} \frac{\partial}{\partial y_k}$$ Only the first component is zero. It is rather the co-vector (differential form) $dy_1$ for which $f^* dy_1=df_1=0$ (at $p$). This implies that $f_*$ can not be onto, so it must have a kernel, although you can't tell from the present information how it looks like.