---
tag: 065J822CKN
kind: statement
role: proposition
name: Inclusion of a subspace is continuous
msc: [54B05]
uses: [0D5YHQGDJY, 0SY9ECQ19Q, 0FGCFY1AZX]
sources:
  - url: https://en.wikipedia.org/wiki/Subspace_topology
    locator: Properties
---

## Statement

Let $X$ be a [[topological space|0CNQXVBHCX]], let $S \subseteq X$ have the [[subspace topology|0D5YHQGDJY]] and let $\iota \colon S \to X$ be the inclusion map $\iota(s) = s$.
Then $\iota$ is [[continuous|0SY9ECQ19Q]].

## Remarks

Proof: for $U$ open in $X$, the [[preimage|0FGCFY1AZX]] is $\iota^{-1}(U) = S \cap U$, which is open in $S$.
Thus the [[restriction|0D2MXBV21T]] $f|_S = f \circ \iota$ of a continuous map $f \colon X \to Y$ is continuous.
A map $g \colon Z \to S$ is continuous if and only if $\iota \circ g \colon Z \to X$ is continuous. This is the universal property of the subspace topology.

## Explanations

### L1 human

Putting a subspace back into the big space is a continuous map.
So when you restrict a continuous function to a subset, you get a continuous function.