Finite graph homotopically equivalent to wedge sum of finite circle. Show that if X is a finite graph i.e. a graph with finitely many vertices and finitely many
edges, then X is homotopically equivalent to wedge sum of finitely circles.
I know that I will be able to make the connected components of finite graph homotopically equivalents to wedge sum of finite circles as I can shrink few edges but how do I proceed further
 A: Similar idea as in Eetu Koskela's answer.
Just want to point out that one can refer to Example 1.22 and section 1.B in Hatcher's Algebraic Topology.
First look for a spanning tree of the connected graph. Then use Seifert-van Kampen theorem (where we deformation retract things), we can see that the fundamental group of the graph is a free group, say $F_n$. Take the wedge sum of $n$ circles, and its fundamental group is also $F_n$. Since both our graph and the wedge sum of finite circles (essentially also a connected graph, with loops) are CW complex $K(G,1)$ that have the same fundamental group, they are homotopy equivalent.
A: Intuition goes something like this:
Let $X$ be a finite connected graph. Every tree is contractible, (deformation retractable to a point), therefore if $X$ is a tree then $X \simeq 0$.
Otherwise, consider chordless cycles of $X$, (cycles that don't have subcycles). Any chordless cycle is homotopy equivalent to $S^1$.
Edge is a tree, so for any two chordless cycles sharing an edge, the edge contracts to a point, resulting in homotopy equivalency to $S^1 \vee S^1$. Same goes for $n$ cycles. Therefore, everything besides chordless cycles contracts to a point $x_0$ at which all of $X$'s chordless cycles are wedged at. 
Concatenation of trees and chordless cycles makes up any finite graph. 
Therefore, any finite graph $X$ with $n$ chordless cycles:
$$X \simeq \underbrace{S^1 \vee S^1 \vee \cdots \vee S^1}_{n\text{ times}}$$
