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Why is it that if $f$ is an immersion at $x$, then it is also an immersion on a nonempty open set $U \ni x?$ I have a feeling that one must use the Inverse Function Theorem to prove this, but I'm stumped.

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Let $f:R^n\rightarrow R^m$ be a differentiable map, $f$ is an immersion at $x$ if and only if the rank of $df_x$ is $n$ this is characterized by the fact that the determinant of a $n\times n$ sub-matrix $A_x$ of $df_x$ is not zero. This is also true in a neighborhood of $x$ since $x\rightarrow det(A_x)$ is continue.

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    I don't quite understand the justification for the last line - could you expand on that a bit? – 2017-02-26
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    If $g(x)=det(A_x)\neq 0$, there exists an open interval $U$ wich contains $det(A_x)$ and does not contains $0$, $g^{-1}(U)$ is open since $g$ is continuous. – 2017-02-26
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    Yes, that is basically the Inverse Function Theorem. I don't see why this guarantees injectivity will be preserved on that specific open set. Is it because an invertible linear map is bijective? – 2017-02-26
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    but the determinant of $g=det(A_x)$ will be non zero on an open subset and this means that the rank of $def_y, y\in g^{-1}(U)$ is $n$. – 2017-02-26