Why is Every Unit Speed Curve A Riemannian Immersion I'm trying to convince myself that every unit speed curve $c:\mathbb{R} \to \mathbb{R}^{2}$ is a Riemannian immersion. The differential $c_{\ast,t_{0}}:T_{t_{0}}\mathbb{R} \cong \mathbb{R} \to T_{c(t_{0})}\mathbb{R}^{2} \cong \mathbb{R}^{2}$ is given by $c_{\ast,t_{0}}(t) = (\dot{c^{x}}(t),\dot{c^{y}}(t))$ where $\dot{c^{x}}(t)$ and $\dot{c^{y}}(t)$ are the $x$ and $y$ components of $\dot{c}$ at $t$. Clearly $c_{\ast,t_{0}}(t) = 0$ if and only if $$\dot{c^{x}}(t) = \dot{c^{y}}(t) = 0$$ which never occurs because $|\dot{c}(t)| = 1$ for all $t$. So $c$ is an immersion. What I can't show is that it's actually a Riemannian immersion. Indeed, $g_{\mathbb{R}}(t,t') = tt'$ while $c^{\ast}g_{\mathbb{R}^{2}}(t,t') = g_{\mathbb{R}^{2}}(\dot{c^{x}}(t),\dot{c^{y}}(t))(\dot{c^{x}}(t'),\dot{c^{y}}(t')) = \dot{c^{x}}(t)\dot{c^{x}}(t')+\dot{c^{y}}(t)\dot{c^{y}}(t')$ and this certially doesn't have to be $tt'$ right? Or am I missing something here?
 A: 
Indeed, $g_{\mathbb{R}}(t,t') = tt'$ while
  $c^{\ast}g_{\mathbb{R}^{2}}(t,t') =
> g_{\mathbb{R}^{2}}(\dot{c^{x}}(t),\dot{c^{y}}(t))(\dot{c^{x}}(t'),\dot{c^{y}}(t'))
> = \dot{c^{x}}(t)\dot{c^{x}}(t')+\dot{c^{y}}(t)\dot{c^{y}}(t')$ and this certially doesn't have to be $tt'$ right? Or am I missing
  something here?

This last part is essentially completely wrong. But you're not far from the right answer despite that. 
Let's pick a point $b \in \Bbb R$. Then you need to compare $g_R(b)$ (a bilinear form on $T_b \Bbb R$) with $g_{R^2} c(b)$ -- a different bilinear form. 
I'm going to call the first form $h$, and the second one $k$, to avoid rococo notation. What you need to prove is that for any two vectors $u,v$ in $T_b \Bbb R$, we have
$$
h[u, v] = k[ u', v' ]
$$
where $u' = dc(b)[u], v'= dc(b)[v]$. 
Now as you've observed, we can identify $T_b \Bbb R$ with $\Bbb R$, so $u$ and $v$ are just numbers. Second, we can identify $T_{c(b)} \Bbb R^2$ with $R^2$. And the linear map $dc(b)$ from the first to the second becomes
$$
s \mapsto s \pmatrix{c_x'(b)\\ c_y'(b)}
$$
Now let's write out the two things we want to show are equal. The first is
$$
h[u, v] = uv. 
$$
The second is 
$$
k(u', v') = \left(u\pmatrix{c_x'(b)\\ c_y'(b)}\right) \cdot \left(v \pmatrix{c_x'(b)\\ c_y'(b)}\right) = uv \left( \pmatrix{c_x'(b)\\ c_y'(b)}\cdot \pmatrix{c_x'(b)\\ c_y'(b)} \right).
$$
This last inner product is just the squared length of the tangent vector to $c$ at the point $b$, which is $1$, so we get that $k(u',v') = uv$ as well, and we're done. 
