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I was reading the an article which made the following statement:

since $g$ is a continuous function which is bounded from below by a positive number, it has a continuous inverse?

The only other relevant information about $g$ is that it is a function from a compact set to $\mathbb{R}$, but that is used in the statement that is attains its minimum. My question is:

Why exactly can one conclude $g$ has a continuous inverse?

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up vote 4 down vote accepted

They’re not talking about an inverse function, but rather about a multiplicative inverse in the algebra of continuous functions on $X$. The inverse in question is the function $h:X\to\Bbb R$ that takes $x$ to $\frac1{g(x)}$. This is clearly continuous: it’s the composition of the continuous function $g$ with the function $$\Bbb R^+\to\Bbb R^+:x\mapsto\frac1x\;,$$ which is continuous on $\Bbb R^+$. And the product function $gh$ is the constant function $1$ on $X$, which is the multiplicative identity of $C(X)$.

In other words, $h$ is the inverse of $g$ in the sense that $gh=hg=1_X$, where the product is the usual pointwise product, and $1_X(x)=1$ for each $x\in X$; it’s not the inverse in the sense that $h\circ g=\text{id}_X$, the identity function on $X$.

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