Answer:

Suppose f:G₁->G₂ is a graph isomorphism, and G₁ is bipartite. We need to show that G₂ is bipartite.

Let V₁ and V₂ be the vertex sets of G₁ and G₂. G₁ is bipartite means we can partition vertex set V₁ into two disjoint sets X and Y, such that V₁=XUY, and such that every edge of G₁ connects a vertex x of X with a vertex y of Y.

Now consider the subsets f(X) and f(Y) of V₂. Since f is a bijection from V₁ to V₂, f(X) and f(Y) are disjoint and f(X)Uf(Y)=V₂.

Now consider two vertices w,z of V₂. If w and z both lie in f(X), then f^-1(w) and f^-1(z) both lie in X. Since G₁ is bipartite, there is no edge between f^-1(w) and f^-1(z), since f preserves adjacencies, there is no edge between w and z in G₂. Similarly, if w and z both lie in f(Y), there is no edge between them.

Every edge of G₂ connects a vertex of f(X) with a vertex of f(Y), so G₂ is bipartite, with bipartition (f(X),f(Y)).