Metamath Proof Explorer


Theorem pcoptcl

Description: A constant function is a path from Y to itself. (Contributed by Mario Carneiro, 12-Feb-2015) (Revised by Mario Carneiro, 19-Mar-2015)

Ref Expression
Hypothesis pcopt.1 𝑃 = ( ( 0 [,] 1 ) × { 𝑌 } )
Assertion pcoptcl ( ( 𝐽 ∈ ( TopOn ‘ 𝑋 ) ∧ 𝑌𝑋 ) → ( 𝑃 ∈ ( II Cn 𝐽 ) ∧ ( 𝑃 ‘ 0 ) = 𝑌 ∧ ( 𝑃 ‘ 1 ) = 𝑌 ) )

Proof

Step Hyp Ref Expression
1 pcopt.1 𝑃 = ( ( 0 [,] 1 ) × { 𝑌 } )
2 iitopon II ∈ ( TopOn ‘ ( 0 [,] 1 ) )
3 cnconst2 ( ( II ∈ ( TopOn ‘ ( 0 [,] 1 ) ) ∧ 𝐽 ∈ ( TopOn ‘ 𝑋 ) ∧ 𝑌𝑋 ) → ( ( 0 [,] 1 ) × { 𝑌 } ) ∈ ( II Cn 𝐽 ) )
4 2 3 mp3an1 ( ( 𝐽 ∈ ( TopOn ‘ 𝑋 ) ∧ 𝑌𝑋 ) → ( ( 0 [,] 1 ) × { 𝑌 } ) ∈ ( II Cn 𝐽 ) )
5 1 4 eqeltrid ( ( 𝐽 ∈ ( TopOn ‘ 𝑋 ) ∧ 𝑌𝑋 ) → 𝑃 ∈ ( II Cn 𝐽 ) )
6 1 fveq1i ( 𝑃 ‘ 0 ) = ( ( ( 0 [,] 1 ) × { 𝑌 } ) ‘ 0 )
7 simpr ( ( 𝐽 ∈ ( TopOn ‘ 𝑋 ) ∧ 𝑌𝑋 ) → 𝑌𝑋 )
8 0elunit 0 ∈ ( 0 [,] 1 )
9 fvconst2g ( ( 𝑌𝑋 ∧ 0 ∈ ( 0 [,] 1 ) ) → ( ( ( 0 [,] 1 ) × { 𝑌 } ) ‘ 0 ) = 𝑌 )
10 7 8 9 sylancl ( ( 𝐽 ∈ ( TopOn ‘ 𝑋 ) ∧ 𝑌𝑋 ) → ( ( ( 0 [,] 1 ) × { 𝑌 } ) ‘ 0 ) = 𝑌 )
11 6 10 eqtrid ( ( 𝐽 ∈ ( TopOn ‘ 𝑋 ) ∧ 𝑌𝑋 ) → ( 𝑃 ‘ 0 ) = 𝑌 )
12 1 fveq1i ( 𝑃 ‘ 1 ) = ( ( ( 0 [,] 1 ) × { 𝑌 } ) ‘ 1 )
13 1elunit 1 ∈ ( 0 [,] 1 )
14 fvconst2g ( ( 𝑌𝑋 ∧ 1 ∈ ( 0 [,] 1 ) ) → ( ( ( 0 [,] 1 ) × { 𝑌 } ) ‘ 1 ) = 𝑌 )
15 7 13 14 sylancl ( ( 𝐽 ∈ ( TopOn ‘ 𝑋 ) ∧ 𝑌𝑋 ) → ( ( ( 0 [,] 1 ) × { 𝑌 } ) ‘ 1 ) = 𝑌 )
16 12 15 eqtrid ( ( 𝐽 ∈ ( TopOn ‘ 𝑋 ) ∧ 𝑌𝑋 ) → ( 𝑃 ‘ 1 ) = 𝑌 )
17 5 11 16 3jca ( ( 𝐽 ∈ ( TopOn ‘ 𝑋 ) ∧ 𝑌𝑋 ) → ( 𝑃 ∈ ( II Cn 𝐽 ) ∧ ( 𝑃 ‘ 0 ) = 𝑌 ∧ ( 𝑃 ‘ 1 ) = 𝑌 ) )