An application of Tube lemma There is a problem :If X and Y are two topological spaces with Y is compact then f:X×Y->X is closed map is an application of Tube lemma;But I have a simple approach. The topology defined power set of X obviously has the closed set X;similarly Y is also closed.So X×Y is also Closed and it maps to closed set X.Hence f is closed mapping.But I didn't the use the fact that Y is compact; My friend sujjest me to study Tube lemma...But still I failed to understand the application of that lemma in my problem.Did I make any mistakes in my approach?
 A: It doesn't work if $Y$ is not compact, for example, consider $X = Y = \Bbb R$ and the hyperbola $H = \{(x,y) \in \Bbb R^2 \mid xy = 1\}$. Then, $f(H) = \{x \in \Bbb R : x \neq 0\}$ which is not closed. 
Your mistake is that $f$ has to send arbitrary closed set to closed set, not just $X$. 
Here is the stament of the Tube lemma I know (c.f Lee, Introduction to Topological manifolds, page 95) : 
Let $X$ a topological space and $Y$ a compact topological space. If $x \in X$ and $U \subset X \times Y$ contains $x \times Y$ there is $V \subset X$ a neighborhood of $x$ with $V \times Y \subset U$.
Now here is the proof assuming the Tube lemma : We want to show that $f : X \times Y \to X, (x,y) \mapsto x$ is closed, assuming $Y$ is compact. Let $C$ a closed set in $X \times Y$, and $\tilde x \in X \backslash f(C)$. We take $U = f^{-1}(X \backslash C)$, by definition $\tilde x \times Y \subset f^{-1}(X \backslash C)$. So there is a $V \subset X$, neighborhood of $\tilde x$ with $V \times Y \subset U$ by the tube lemma, which means exactly that $f(V \times Y) \cap f(C) = \emptyset$, i.e $V$ is an open neighborhood of $\tilde x$ which does not intersect $f(C)$ and it follows that $f(C)$ is closed. 
A: It does send the closed set $X\times Y$ to the closed set $X$, yes. However, just because it works for one closed set does not mean it works for them all, so you haven't shown that $f$ is a closed mapping.
