Is a sub manifold of Euclidean space with the same tangent space at each point an affine subspace?

Question

I have a smooth (not necessarily connected) submanifold $M \subset R^n$ of dimension $m$ with the property that the tangent space to $M$ at each point $p$ is "the same". Now I understand that in general it makes no sense to talk about equality of tangent spaces (just about isomorphisms) but in this Euclidean case by "same" I mean that if I translate each $T_p M$ to the origin by subtracting the vector $p$ then it's the same fixed $m$-dimensional subspace $H$ of $R^n$, independent of $p$. Here I'm using of course the usual global coordinates on $R^n$. If I'm wrong then please correct me.

Question: Does it follow that then $M$ is "parallel" to $H$? That is, that $M$ must be contained in a translate of $H$? I think this is trivially true because if $M$ is contained in a translate of $H$ then its tangent space at each point meets the conditions I gave in the first paragraph. Thus, it "must" be $M$. I'd just like to know if my thought process is correct or if I've gone off the rails somewhere. Thank You.

Edit: Changed title to accurately reflect question, thanks to Ted Shifrin.

Edit: Again responding to a comment by Ted Shifrin, who pointed out that I proved necessity rather than the desired sufficiency. My idea was to show that if you have a manifold with the "same" tangent plane at each point then it is an affine subspace, much like if you have a curve in $R^2$ with the same derivative everywhere it's a line. So, I first specified that the tangent spaces at each point of $M$ must be the same, then observed that an affine subspace has that property therefore, by "uniqueness" (granted, I'm not quoting a specific result), $M$ must be (contained in) such a subspace. That was my thought process anyway, I realize it's a simple result but I don't want to make any mistakes.

Answer 1

Following a suggestion by Ted Shifrin: We can view $M$ as the graph of a smooth function $f: U \rightarrow H^\perp$, where $U \subset H$ is some connected open set. If we choose coordinates so that $H = (x_1,\ldots,x_m)$ we then have $$ M = (x_1,\ldots,x_m,f^1(x_1,\ldots,x_m),\ldots,f^{n-m}(x_1,\ldots,x_m)) $$ for $(x_1,\ldots,x_m) \in U$. At each $p \in M$ the tangent space $T_pM$ is spanned by the $m$ vectors $$ (1,0,\ldots,0,f^1_1(p),\ldots,f^{n-m}_1(p))\ldots (0,\ldots,1,f^1_{m}(p),\ldots,f^{n-m}_m(p));$$ this would be consistent with my original scenario if I had specified that $H$ was the $(x_1,\ldots,x_m,0,\ldots,0)$ plane. In that case we must have $f^i_j(p) = 0,\; 1 \leq i \leq n-m,\: 1 \leq j \leq m$. In other words, each $f^i(p)$ must be a constant function, call it $k_{i}$; therefore $$ M = (x_1,\ldots,x_m,k_1,\ldots,k_{n-m}), x \in U$$ which, is an affine subspace (I hope I'm right here).

Now I have made the simplifying assumption that $U$ is open and connected. In particular, as Paul Frost has pointed out, if $M$ is not assumed connected then this doesn't quite work. Actually, this question comes out of some research in which eventually I will just need all this to be true locally, so that's OK.

My thanks to all who commented (and perhaps will comment), I hope I have this right but my math skills are unfortunately not what I'd like them to be, this resource is invaluable.

Update: After thinking about Moishe Kohan's comment, I realize a couple of things. First, if $M$ is disconnected then, as Paul Frost pointed out, this isn't going to work (which is consistent with the idea that in general a manifold is only locally a graph). There are also results (e.g. Proposition 6.32 in Lee's ISM) that give conditions under which a manifold is globally a graph. I am going to do research to discover the "bit more work" Moishe Kohan refers to: At this point I can only see two options. One is to apply a result like Lee's referred to above because I believe the assumption that $T_pM$ is the same at all points of $M$ implies that $\{p\} \times R^{n-m}$ can only intersect $M$ at one point. The other is to try to show that the graph function defining the manifold locally can always be extended because the derivatives are bounded, sort of like when dealing with the maximal domain of integral curves. I hope I am on the right track but will see and again appreciate the help that was offered.

Final Attempt at Proof: Taking into account the comments of the contributors noted above I have come up with the following statement and "proof". Statement: Suppose $M \subset R^n$ is a smooth, connected submanifold of dimension $m$. Further, suppose that at each $p \in M$ the tangent space $T_p(M) = H$, where $H \subset R^n$ is a fixed linear subspace of dimension $m$. Then $M \subset H + \{p_0\}$, where $p_0$ is any point of $M$; i.e. $M$ is contained in a translate of $H$. Proof: Fix $p_0 \in M$ and let $p^\prime \in M$ be any other point; since connected manifolds are path connected (at least in Euclidean space) there exists a continuous curve $\gamma: [0,t] \rightarrow M$ connecting $p_0$ and $p^\prime$. This curve can actually be made to be piecewise smooth (here I'm drawing on posts seen on MSE), say $\gamma: [t_0 = 0, t_1,t_2,\ldots,t_{n-1},t_n = t] \rightarrow M$, with $\gamma(t_0) = p_0,\; \gamma(t_n) = p^\prime$. But at every point $t$ at which $\gamma$ is differentiable we must have $\gamma^\prime (t) \in T_{\gamma (t)}M = H$ (we use one-sided derivatives at the endpoints). So $\gamma[t_0,t_1) \in \gamma(t_0) + H = H + \{p_0\}$; furthermore, since $H + \{p_0\}$ is closed and $\gamma$ obviously continuous, $\gamma(t_1) \in H + \{p_0\}$. Continuing in this fashion we conclude that $\gamma(t_n) = p^\prime \in H + \{p_0\}$. Noting that $p^\prime$ was arbitrary completes the proof.

As I say, this is my best effort. I hope I have gotten it "mostly" right but, as always, will take whatever criticism is leveled at me.

Sorry, one more thing... This is taking on a life of its own. I am aware that just because the tangent to a curve is everywhere contained in a hyperplane does not in general mean the curve "stays" in that hyperplane. I have had occasion to make use of Whitney's example of a curve not constant on a connected set of critical points; the difference is that in the example the domain of the function is an unrectifiable curve in $R^2$ while here the curve is the image of a subinterval of $R$ by a smooth (almost everywhere) function. So I waved my hands a bit on that one.