# The graph of a smooth map is a manifold

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...

• What does the inverse function theorem tell you? – Michael Burr Jan 16 '18 at 3:49
• @MichaelBurr: in the book, the theorem is introduced after the problem is given. – user500094 Jan 16 '18 at 3:55

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.
• But why is the projection smooth? I think I'm stuck at this proof exactly at the same point as in my question. This map $(x,f(x))\mapsto x$ cannot be extended to a smooth map $R^k\times R^k\rightarrow R^k$ by the same formula (like I did in the preceding points in my question) since $f(X)\ne R^k$. – user500094 Jan 16 '18 at 4:06
• @user500094 I don't understand what you mean about extending to a smooth map... projection is already a smooth map on all of $\mathbb{R}^k\times\mathbb{R}^k$? – ziggurism Jan 16 '18 at 4:09