Here is a problem from Guillemin-Pollack:
The graph of a map $f: X\rightarrow Y$ is the subset of $X\times Y$ defined by $$graph(f)=\{(x,f(x)):x\in X\}.$$ Define $F: X\rightarrow graph(f)$ by $F(x)=(x,f(x))$. Show that if $f$ is smooth, $F$ is a diffeomorphism; thus $graph(f)$ is a manifold if $X$ is.
(Note that the notion of a manifold in this book is defined as follows: a subset of $R^k$ is a manifold of dimension $n$ if every points admits a neighborhood diffeomorphic to an open subset of $R^n$. For the definition of a smooth map see this question.)
Assume $X \subset R^k$ (as remarked above, only subsets of $R^k$ are considered in the book). My idea is to regard $F$ as the composition of $X\rightarrow X\times X \rightarrow graph(f), x\mapsto (x,x)\mapsto (x,f(x))$. The first map $x\mapsto (x,x)$ is smooth because it extends by the same formula to a smooth map $R^k\rightarrow R^k\times R^k$ and agrees with the original map on $X$. The second map $(x,x)\mapsto (x,f(x))$ is smooth by The smoothness of a product map.
Further, the first map is clearly bijective and its inverse, $(x,x)\mapsto x$, is smooth since it extends by the same formula to a smooth map $R^k\times R^k\rightarrow R^k$. The second map is also bijective with inverse $X\times f(X)\rightarrow X\times X,\ (x,f(x))\mapsto (x,x)$. But why is the inverse smooth? Now we cannot extend it to a smooth map $R^k\times R^k\rightarrow R^k\times R^k$ by the same formula anymore...
The map $(x,f(x))\mapsto (x,x)$ can be viewed as the composition of the projection map $\operatorname{proj}_1\colon(x,f(x))\mapsto x$ and the diagonal map, both of which are smooth.