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Suppose you have a smooth function $\sigma = (\sigma_1,\sigma_2,\sigma_3):\mathbb{R}^2\rightarrow\mathbb{R}^3$. Write $\sigma_x:=\left(\frac{\partial\sigma_1}{\partial x},\frac{\partial\sigma_2}{\partial x},\frac{\partial\sigma_3}{\partial x}\right)$ and similarly for $\sigma_y$. If at some $p\in\mathbb{R}^2$ we have $\sigma_x = \sigma_y$, then is it still possible for $\sigma$ to be a bijection onto its image?

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Sure. For instance, let $\sigma(x,y)=(x^3,y^3,0)$. Then $\sigma_x(0,0)=\sigma_y(0,0)=(0,0,0)$, but $\sigma$ is a smooth injection.

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    Ah I should've mentioned it, I'd like to rule out the case when both are 0. Is it still possible then? – 2017-02-28
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    Sure, just change it to $\sigma(x,y)=(x^3,y^3,x+y)$. – 2017-02-28
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    Thanks! How about if we make $\sigma:\mathbb{R}^2\rightarrow\mathbb{R}^2$? – 2017-02-28
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    Try $\sigma(x,y)=(x^3,x+y)$. – 2017-02-28
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    Maybe slightly more interesting: can one do the same thing if $\sigma$ is (at least around a neighbourhood) a diffeomorphism? This would give a nice counter-example to the converse of the inverse image theorem. – 2017-02-28
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    No: if $\sigma$ is a diffeomorphism (onto its image), the total derivative of $\sigma$ must be injective. – 2017-02-28