Matrix multiplication is jointly continuous

Idea. Product topology on $M_n\times M_n\cong\mathbb{R}^{2n^2}$ and polynomial dependence of the entries give continuity at once. A direct estimate gives more: local Lipschitz control.

Step 1 (qualitative). $(AB)_{ij}=\sum_{k}a_{ik}b_{kj}$ is a polynomial in the $2n^2$ coordinates $(a_{ik},b_{kj})$, so it is continuous (see Problem 66). All $n^2$ entries are continuous, hence so is the map.

Step 2 (quantitative). Use the operator norm, which is submultiplicative. For $(A',B')$ near $(A,B)$:
$$A'B'-AB=(A'-A)B'+A(B'-B),$$
check: $(A'-A)B'+A(B'-B)=A'B'-AB'+AB'-AB=A'B'-AB$. Hence
$$\|A'B'-AB\|\le\|A'-A\|\,\|B'\|+\|A\|\,\|B'-B\|.$$
If $\|B'-B\|\le1$, then $\|B'\|\le\|B\|+1$, so
$$\|A'B'-AB\|\le(\|B\|+1)\|A'-A\|+\|A\|\,\|B'-B\|\xrightarrow[(A',B')\to(A,B)]{}0.$$

Remark. The map is *not* uniformly continuous on $M_n\times M_n$: take $A=B=tI$ and $A'=B'=(t+\frac1t)I$; the distance is $\frac{2\sqrt n}{t}\to0$ but $\|A'B'-AB\|=\|(2+t^{-2})I\|\ge2$. This is the same phenomenon as for $(x,y)\mapsto xy$ on $\mathbb{R}^2$. Together with the next problems it shows that $GL_n(\mathbb{R})$ is a topological group.