Proposition

Let $X, Y$ be topological spaces.

Every constant map $f: X \rightarrow Y$ is continuous.
The identity map $\text{Id}_X: X \rightarrow X$ is continuous.
If $f: X \rightarrow Y$ is continuous, so is the restriction of $f$ to any open subset of $X$.

Solution

1

Let $f: X \rightarrow Y$ be a constant map. Let $y$ be the constant to which $f$ maps every element of $X$. Let $V \subset Y$ be an open subset.

If $y \in V$, then $f^{-1}(V) = X$, and $X$ is open in $X$.
If $y \notin V$, then $f^{-1}(V) = \emptyset$, and $\emptyset$ is open in $X$.

Therefore, $f$ is continuous.

2

Let $f: X \rightarrow X$ be the identity map. Let $U \subset X$ be an open subset. Then $f^{-1}(U) = U$, so $f^{-1}(U)$ is open. Therefore, $f$ is continuous.

3

Let $A$ be any subset of $X$.

Consider $f_A: A \rightarrow f(A)$. Let $U \subset f(A)$ be an open subset. Then $U = V \cap f(A)$ for some $V$ that is open in $Y$.

$f_A^{-1}(U) = f^{-1}(U) \cap A = f^{-1}(V \cap f(A)) \cap A = f^{-1}(V) \cap f^{-1}(f(A)) \cap A$.

$f^{-1}(V)$ is open in $X$ because $f$ is continuous.
$A \subset f^{-1}(f(A))$, so $f^{-1}(f(A)) \cap A = A$.

Therefore, $f_A^{-1}(U) = f^{-1}(V) \cap A$, so it is open in $A$. Hence, $f_A$ is continuous.