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Let $n$ be a natural number, and let $m \colon= n^2$. Let $\mathrm{GL}(n)$ denote the set of all the non-singular $n \times n$ matrices of real numbers. Each matrix of order $n \times n$ has $n^2 = m$ real entries and so can be regarded as an element of the Euclidean space $\mathbb{R}^m$.

Now let us give $\mathbb{R}^m$ the topology having as a basis the collection of all possible Cartesian products of the form $$ \prod_{i=1}^m \left( a_i, b_i \right), $$ where, for each $i = 1, \ldots, m$, $a_i$ and $b_i$ are real numbers such that $a_i < b_i$.

Let us consider $\mathrm{GL}(n)$ to be a (topological) subspace of $\mathbb{R}^m$.

Let the maps $f \colon \mathrm{GL}(n) \times \mathrm{GL}(n) \to \mathrm{GL}(n)$ and $g \colon \mathrm{GL}(n) \to \mathrm{GL}(n)$ be defined by $$ f( A \times B) \colon= AB \qquad \mbox{ for all } A \times B \in \mathrm{GL}(n) \times \mathrm{GL}(n), $$ and $$ g(A) \colon= A^{-1} \qquad \mbox{ for all } A \in \mathrm{GL}(n). $$

Then how to --- rigorously and explicitly --- show the maps $f$ and $g$ to be continuous? I would be grateful for a proof using only the results in Secs. 18 through 21 of Munkres' Topology, 2nd edition.

I know that the addition, subtraction, and multiplication operations are continuous maps of $\mathbb{R} \times \mathbb{R}$ into $\mathbb{R}$; and the division operation is a continuous map of $\mathbb{R} \times \left( \mathbb{R} \setminus \{ \ 0 \ \} \right)$ into $\mathbb{R}$.

And, I know that the projection of $\mathbb{R}^k$ onto any coordinate is a continuous map.

Of course, the sum, difference, product, quotient (whenever it is defined) of continuous complex-valued maps are continuous. And, the composite map (whenever it is defined) of (two or mmore) continuous maps is continuous.

But, I'm finding it difficult to decide which facts to call to my aid!!

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    $\begingroup$ You're pretty much saying it. Everything is a polynomial or rational function of the entries. $\endgroup$ – Randall Nov 13 '17 at 16:07
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$\textbf{Hint :}$

The multiplication is just polynomial of the entries so its continous. By Cramers rule and continuity of $det : GL(n) \rightarrow \mathbb{R}$ (because it is also a polynomial), the inverse is continous.

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    $\begingroup$ A commentor's gloss: being able to show that $A\mapsto A^{-1}$ is continuous (or better, that the entries of $A^{-1}$ are rational functions of entries of $A$) is basically the purpose of Cramer's rule. It is easy to miss this in a linear algebra course. $\endgroup$ – Kyle Miller Nov 13 '17 at 18:55
  • $\begingroup$ @KyleMiller thank you for your comment. I've just visited the link to your profile page here on Math SE, but I'm unable to access the materials from the UC Berkeley, which you've shared on your profile page. What are my options? $\endgroup$ – Saaqib Mahmood Nov 13 '17 at 20:42
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    $\begingroup$ @SaaqibMahmuud Sorry, I handwrote the lecture notes, where pages 4-7 of math.berkeley.edu/~kmill/math54su16/lecnotes_06_30.pdf have a basic introduction to Cramer's rule. A higher-level overview is that the product of a matrix and its adjugate is the determinant times the identity, where the adjugate is formed from the minors of the matrix. This implies the inverse of the matrix is the adjugate divided by the determinant. The determinant is a polynomial function of the entries of a matrix, so is continuous. Because of this, invertible matrices have neighborhoods of invertible matrices. $\endgroup$ – Kyle Miller Nov 13 '17 at 22:12

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