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Let $(M,d)$ and $(N,q)$ be metric spaces.

The operator $T:M\longrightarrow N$ is contractive in the sense that $q(T(m_1),T(m_2)) \leq c d(m_1, m_2)$ for some $c\in [0,1)$.

Similarly, the operator $J:N\longrightarrow M$ is contractive in the "opposite direction", so that $d(J(n_1),J(n_2)) \leq k q(n_1, n_2)$ for some $k\in[0,1)$.

Is it necessarily true that the compositions of these $(J\circ T): M\longrightarrow M$ is also a contraction mapping?

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1 Answer 1

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Yes: $d(J(T(m_1)), J(T(m_2))) \leq kq(T(m_1), T(m_2)) \leq kcd(m_1, m_2)$.

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